INTERNATIONAL JOURNAL OF GEOMETRY Vol. 10 (2021), No. 3, 21 - 47

MORE NEW CHARACTERIZATIONS OF TANGENTIAL

MARTIN JOSEFSSON and MARIO DALC´IN

Abstract. This is a continuation to the investigations we made in [15] about tangential quadrilaterals. Here we prove 24 more necessary and suf- ficient conditions for when a convex can have an incircle.

1. Introduction A convex quadrilateral in which a can be drawn that is internally to all four sides is most often called a or a circumscribed quadrilateral, but several other similar names have been used (see [6, p. 65]). This type of quadrilateral was first considered in the thirteenth century by Jordanus Nemorarius according to [1, p. 300], but it is unclear what discoveries he made. The most important and fundamental theorem for a tangential quadrilateral ABCD is called Pitot’s theorem after the French engineer Henri Pitot, who in 1725 proved that (1) AB + CD = BC + DA. This condition is also sufficient, which was proved by J. B. Durrande in 1815 [18, p. 64]. At the time of writing this paper, more than ten different proofs of this converse are known (most of these references are given in [14]). Just over a decade ago, the interest in these quadrilaterals was greatly increased, resulting in a large number of necessary and sufficient conditions being discovered and proved. The papers [16, 6, 8, 9, 10, 15] contain al- together 49 different characterizations (either proved or reviewed), and the present paper is a continuation to the latter of these. Three other were proved in [2], [5, p. 395] and [12, p. 148]. Adding the 24 necessary and sufficient conditions in this paper results in a total of 76, and makes the tangential quadrilateral the undisputed leader on the list of objects in ge- ometry with the greatest number of known characterizations (the is on second place with 47 published characterizations). ————————————– Keywords and phrases: Tangential quadrilateral, Pitot’s theorem, cyclic quadrilateral, , Newton’s (2020)Mathematics Subject Classification: 51M04 Received: 07.10.2020. In revised form: 06.11.2020. Accepted: 02.11.2020. 22 Martin Josefsson and Mario Dalc´ın

2. Variations on an old theorem In this section we shall study three reformulations of Pitot’s theorem. We call a circle tangent to one side of a quadrilateral and the extensions of the adjacent two sides an escribed circle, as was done in [8, p. 71]. Theorem 2.1. In a convex quadrilateral, the sum of the four distances along the extensions of opposite sides between each and the tangency of the adjacent escribed circle are equal for both pairs of opposite sides if and only if it is a tangential quadrilateral.

Figure 1. ABCD is tangential ⇔ equality (2) holds

Proof. Using notations as in Figure 1, we have according to the two tangent theorem that AE = AQ, EB = BL and so on. By Pitot’s theorem, ABCD is tangential if and only if AE + EB + CG + GD = BF + FC + DH + HA which is equivalent to (2) AQ + BL + CM + DP = BK + CN + DO + AJ. This last equality is what the theorem states. 

The following characterization is related to Theorem 3.2 in [15]. Theorem 2.2. In a convex quadrilateral there are four excircles to the four overlapping created by the that are tangent to the diago- nals. The sum of the four distances between the vertices and the tangency points on the extensions of two opposite sides are equal to the sum of the four distances between the vertices and the tangency points on the extensions of the other pair of opposite sides if and only if it is a tangential quadrilateral. Proof. In [15, p. 57], we proved that in all convex quadrilaterals, ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ BC +DA−CD −AB = 2(S Z −X Y )= S Z +V W −X Y −U T More new characterizations of tangential quadrilaterals 23

Figure 2. ABCD is tangential ⇔ equality (3) holds where the last equality is due to the fact that S′′Z′′ = V ′′W ′′ and X′′Y ′′ = U ′′T ′′, also according to [15, p. 57] (see Figure 2). Thus we get ′′ ′′ ′′ ′′ BC + DA − CD − AB = S A + AB + BZ + V D + DC + CW − X′′A − AD − DY ′′ − U ′′B − BC − CT ′′ which is simplified as ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ 2(BC+DA−CD−AB)= S A+BZ +V D+CW −X A−DY −U B−CT . Hence BC + DA = CD + AB is equivalent to ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ (3) AS + BZ + CW + DV = AX + BU + CT + DY which by Pitot’s theorem finishes the proof. 

Next we have a generalization of Theorem 2 in [6]. Two special cases of the ‘if’ part of this characterization were proved as Theorems 5.4.1 and 5.4.2 in [17, p. 150]. Theorem 2.3. Related to a convex quadrilateral ABCD, there is an arbi- trary circle tangent to the lines AB and BC at S and T , an arbitrary circle tangent to the lines BC and CD at U and V , an arbitrary circle tangent to the lines CD and DA at W and X, and an arbitrary circle tangent to the lines DA and AB at Y and Z respectively. The relation ± ZS ± VW = ± TU ± XY holds if and only if it is a tangential quadrilateral, where the minus sign for each term is chosen if the tangency points have reversed alphabetical order on each line when tracing around the quadrilateral in counter clockwise direction starting at S, given that ABCD is labeled in counter clockwise direction. Proof. There are several possible cases with internal, external, or partially external . We start with the simple case in Figure 3. In a convex 24 Martin Josefsson and Mario Dalc´ın

Figure 3. ABCD is tangential ⇔ ZS + VW = T U + XY quadrilateral ABCD with only internal circles, we have AB − BC + CD − DA = AZ + ZS + SB − BT − TU − UC + CV + VW + WD − DX − XY − Y A = ZS − TU + VW − XY according to the two tangent theorem (AZ = Y A and so on). The quadri- lateral is tangential if and only if AB +CD = BC +DA, which is equivalent to (4) ZS + VW = TU + XY.

Figure 4. In this case, ABCD is tangential ⇔ ZS + VW = −T U + XY

In a more general case as the one in Figure 4, it holds that AB − BC + CD − DA = − AZ ± ZS + SB − (BT ± TU + UC) + CV ± VW + WD − (DX ± XY − Y A) = ± ZS ∓ TU ± VW ∓ XY where the plus signs are chosen when the tangency points on a (possibly extended) side of the quadrilateral are in alphabetical order, otherwise minus More new characterizations of tangential quadrilaterals 25 signs are chosen. Also, the tangent lengths at an external circle are negative (AZ and Y A in Figure 4). ABCD is a tangential quadrilateral if and only if AB + CD = BC + DA ⇔ ± ZS ± VW = ±TU ± XY. In Figure 5, we have drawn an example where all tangency points are in reversed alphabetical order, making all four terms negative. By shrinking the smallest circle, we can easily get the case when there are two negative terms. We let the reader draw the case with three negative terms. 

Figure 5. ABCD is tangential ⇔ −ZS − VW = −T U − XY

It may seem amazing that there is no size constraint on the four circles. But an increase or decrease of any circle add or subtract the same length to both sides of equality (4).

3. Other subtriangle lengths ′ ′ ′ ′ First we have a variant of the true characterization T1T3 = T2T4 of tan- gential quadrilaterals that appeared in the proof of Theorem 3 in [6]. (Note ′ ′ ′ ′ that the equality T1T2 = T3T4 that was stated in that theorem is not a valid characterization of tangential quadrilaterals, since it is also true in paral- lelograms. The first equality however is case-independent and thus remains true no matter where the subtriangle incircles are tangent to the diagonals, which the second equality is not.) Theorem 3.1. In a convex quadrilateral ABCD where the diagonals in- tersect at P , the distance between the points where two opposite incircles in triangles ABP , BCP , CDP , DAP are tangent to one is equal to the distance between the tangency points for the other two opposite incircles on the other diagonal if and only if ABCD is a tangential quadrilateral. 26 Martin Josefsson and Mario Dalc´ın

Figure 6. ABCD is tangential ⇔ T1X1 = V1Z1

Proof. In all convex quadrilaterals, it holds that U1P = V1P and Y1P = Z1P according to the two tangent theorem; whence U1Y1 = V1Z1 (see Fig- ure 6). In the proof of Theorem 3 in [6], it was proved that ABCD is tangential if and only if T1X1 = U1Y1 (but the origin of that theorem was a solution to a Russian geometry problem by I. Vaynshtejn, as noted in [6, p. 68]). Thus we get that T1X1 = V1Z1 is a characterization of tangential quadrilaterals. 

As a consequence we have the following theorem, where we use the nota- tion [S1T1W1X1] for the of quadrilateral S1T1W1X1. Theorem 3.2. The four tangency points on the diagonals of a convex quadrilateral for two opposite non-overlapping subtriangle incircles are the vertices of an isosceles . The two isosceles created this way have equal area if and only if the original quadrilateral is tangential.

Figure 7. ABCD is tangential ⇔ [S1T1W1X1]=[Y1Z1U1V1] More new characterizations of tangential quadrilaterals 27

Proof. We conclude that opposite chords Y1Z1 and V1U1 are parallel by applying the two tangent theorem, the isosceles theorem and equal alternating in quadrilateral Y1Z1U1V1 (see Figure 7). Since the two diagonals in that trapezoid are equal by the two tangent theorem, it is isosceles. In the same way the trapezoid S1T1W1X1 is isosceles. That their [S1T1W1X1] and [Y1Z1U1V1] are equal yields 1 2 1 2 2 (T1X1) sin θ = 2 (V1Z1) sin(π − θ), by a well-known quadrilateral area theorem. This equality is equivalent to T1X1 = V1Z1 and the stated result follows from Theorem 3.1.  ′ ′ ′ ′ Next we have an excircle version of the equation T1T3 = T2T4 related to Vaynshtejn’s problem. Theorem 3.3. In a convex quadrilateral ABCD where the diagonals in- tersect at P , consider the excircles to triangles ABP , BCP , CDP , DAP that are tangent to one side each of the quadrilateral and the extensions of the diagonals. The distance between the tangency points for two opposite excircles on a diagonal is equal to the distance between the tangency points for the other two excircles on the same diagonal if and only if ABCD is a tangential quadrilateral.

Figure 8. ABCD is tangential ⇔ T2X2 = U2Y2

Proof. Let E, G, S2, T2, U2, W2, X2, Y2 be tangency points as indicated in Figure 8. We have that

AB = AE +EB = AS2 +BT2 = PS2 −P A+PT2 −PB = 2PT2 −P A−PB and in the same way

CD = 2PX2 − PC − PD. Thus AB + CD = 2T2X2 − AC − BD and by symmetry, we get

BC + DA = 2U2Y2 − AC − BD. 28 Martin Josefsson and Mario Dalc´ın

Hence

AB + CD − BC − DA = 2(T2X2 − U2Y2) and it follows that ABCD is tangential if and only if T2X2 = U2Y2. 

There is the following variant of the previous theorem. Theorem 3.4. In a convex quadrilateral ABCD where the diagonals inter- sect at P , consider the excircles to triangles ABP , BCP , CDP , DAP that are tangent to one side each of the quadrilateral and the extensions of the diagonals. The distance between the tangency points for two opposite ex- circles on one diagonal is equal to the distance between the tangency points for the other two excircles on the second diagonal if and only if ABCD is a tangential quadrilateral.

Figure 9. ABCD is tangential ⇔ T2X2 = V2Z2

Proof. Using notations as in Figure 9, then U2P = V2P and Y2P = Z2P according to the two tangent theorem. Thus U2Y2 = V2Z2 holds in all convex quadrilaterals. Hence ABCD is tangential if and only if T2X2 = U2Y2 = V2Z2 according to Theorem 3.3. 

We also have an excircle version of Theorem 3.2. Since the proof is iden- tical to the one for that theorem, we omit the proof. This characterization is illustrated in Figure 10, where [S2T2W2X2] denote the area of S2T2W2X2. Theorem 3.5. The four tangency points on the extensions of the diagonals of a convex quadrilateral for two opposite non-overlapping subtriangle ex- circles, that are each tangent to a side of the quadrilateral, are the vertices of an . The two isosceles trapezoids created this way have equal area if and only if the original quadrilateral is tangential. More new characterizations of tangential quadrilaterals 29

Figure 10. ABCD is tangential ⇔ [S2T2W2X2]=[Y2Z2U2V2]

4. bisectors First we prove two basic theorems related to the extensions of a pair of opposite sides. Theorem 4.1. In a convex quadrilateral that is not a trapezoid, where the angle bisectors to two opposite vertex angles intersect at an interior point I, let E and G be the orthogonal projections of I on a pair of opposite sides. The angles between these sides and the transversal EG, on the same side of EG, are equal if and only if it is a tangential quadrilateral.

Figure 11. ABCD is tangential ⇔ ∠GEQ = ∠EGQ

Proof. If ABCD is a tangential quadrilateral, then the incircle is tangent to AB and CD at E and G respectively. Extending these two opposite sides so that they intersect at a point Q, we see that EQ = GQ according to the two tangent theorem, making triangle QEG isosceles. Thus the two angles on one side of EG are equal. Conversely, in a convex quadrilateral with a pair of such equal angles, we get an QEG, so EQ = GQ. Then EI = GI (RHS congruency). By the construction of I, it is equidistant to the pairs of adjacent sides AB, BC and CD, DA (see Figure 11). Since also EI = GI, 30 Martin Josefsson and Mario Dalc´ın we have that I is equidistant to all the sides, making ABCD a tangential quadrilateral (with I). 

It was proved in [10, pp. 401–402] that a convex quadrilateral is tangential if and only if either three internal angle bisectors are concurrent, or the two angle bisectors at the intersections of the extensions of opposite sides and one internal angle bisector are concurrent. A third variant is: Theorem 4.2. In a convex quadrilateral that is not a trapezoid, let a pair of opposite sides intersect at a point Q. The angle bisectors at a pair of opposite vertex angles and the angle bisector at Q are concurrent inside the quadrilateral if and only if it is a tangential quadrilateral.

Figure 12. ABCD is tangential ⇔ QI is an angle bisector at Q

Proof. In a tangential quadrilateral, let I be the incenter and G and E the orthogonal projections of I on a pair of opposite sides. Then GI = EI (the inradius) and GQ = EQ by the two tangent theorem, so triangles GQI and EQI are congruent (SSS), see Figure 12. Then ∠GQI = ∠EQI, making QI the angle bisector at Q, concurrent with two opposite internal vertex angle bisectors. Conversely, suppose that the three stated angle bisectors intersect at a point I in a convex quadrilateral, and that the orthogonal projections of I on the sides are E, F , G, H. Then GI = HI, EI = FI (due to congruent triangles created by the internal angle bisectors) and GI = EI (due to congruent triangles QGI and QEI). Hence HI = GI = EI = FI, so ABCD is a tangential quadrilateral (with incenter I). 

Next we have a characterization related to Theorem 5.2 in [15]. Theorem 4.3. In a convex quadrilateral ABCD, let the incircles in trian- gles ABC, BCD, CDA, DAB be tangent to the sides AB, BC, CD, DA, AB at points S′, T ′, U ′, V ′, W ′, X′, Y ′, Z′. Suppose the lines Z′U ′ and V ′Y ′ intersect at E, and the lines X′S′ and T ′W ′ intersect at F . Then the angle bisectors at E and F are if and only if ABCD is a tangential quadrilateral. Proof. If Z′U ′ intersects X′S′ at K and V ′Y ′ intersects T ′W ′ at M (see Figure 13), then in the proof of Theorem 5.2 in [15] it was proved that ABCD is tangential if and only if it holds that ′ ′ ′ ′ ∠A + ∠B + ∠C + ∠D ∠X KU + ∠Y MT = = π. 2 More new characterizations of tangential quadrilaterals 31

π Figure 13. ABCD is tangential ⇔ ψ = 2

In the proof of Theorem 6.2 in [13] it was proved that the angle ψ between the two considered angle bisectors in a convex quadrilateral is given by ∠X′KU ′ + ∠Y ′MT ′ ψ = . 2 π So when ABCD is tangential, we have ψ = 2 . Conversely, when ABCD is not tangential, then we have ∠X′KU ′ + ∠ ′ ′ π  Y MT =6 π. This means ψ =6 2 , completing the proof.

There is the following excircle version of the previous theorem. Theorem 4.4. In a convex quadrilateral ABCD, let the excircles to trian- gles ABC, BCD, CDA, DAB that are tangent to the diagonals, be tangent to the sides AB, BC, CD, DA, AB at points S′′, T ′′, U ′′, V ′′, W ′′, X′′, Y ′′, Z′′. Suppose the lines Z′′U ′′ and V ′′Y ′′ intersect at E′, and the lines X′′S′′ and T ′′W ′′ intersect at F ′. Then the angle bisectors at E′ and F ′ are perpendicular if and only if ABCD is a tangential quadrilateral. Proof. If S′′X′′ intersects U ′′Z′′ at K′, U ′′Z′′ intersects W ′′T ′′ at L′, W ′′T ′′ intersects Y ′′V ′′ at M ′ and Y ′′V ′′ intersects S′′X′′ at N ′ (see Figure 14), then according to Theorem 5.3 in [15] it holds that ABCD is tangential if and only if K′L′M ′N ′ is cyclic. In the proof of Theorem 6.2 in [13] it was proved that the angle ω between the two considered angle bisectors in a convex quadrilateral is given by ∠X′′K′U ′′ + ∠Y ′′M ′W ′′ ω = . 2 π Hence when ABCD is tangential, ω = 2 . Conversely, when ABCD is not tangential, then ∠X′′K′U ′′+∠Y ′′M ′W ′′ =6 π  π. Thus ω =6 2 , so the angle bisectors are not perpendicular. 32 Martin Josefsson and Mario Dalc´ın

π Figure 14. ABCD is tangential ⇔ ω = 2

5. Cyclic quadrilaterals It is well known that the four angle bisectors to the vertex angles in a tangential quadrilateral are concurrent at the incenter. We start this section with a variant of that result. Theorem 5.1. In a convex quadrilateral ABCD, suppose two opposite angle bisectors intersect at an interior point I. Let E, F , G, H be the orthogonal projections of I on the sides AB, BC, CD, DA respectively. Then EFGH is a cyclic quadrilateral if and only if ABCD is a tangential quadrilateral.

Figure 15. ABCD is tangential ⇔ EFGH is cyclic

Proof. When ABCD is tangential, E, F , G, H are points on the incircle so EFGH is cyclic. Conversely, when EFGH is cyclic, we have that triangles AEI and AHI are congruent (AAS), so EI = HI and AE = AH, and in the same way More new characterizations of tangential quadrilaterals 33

CFI and CGI are congruent, so FI = GI and CF = CG (see Figure 15). Thus AEIH and CFIG are kites, so AI ⊥ EH and CI ⊥ FG, which means that AI and CI are perpendicular bisectors to EH and FG respectively, and these intersect at the center I of the circumcircle to EFGH. Since E, F , G, H are all on this circle, we get EI = FI, which implies that I is equidistant to the sides AB, BC, CD, DA. Hence ABCD is a tangential quadrilateral. 

The previous theorem might appear trivial, but when proving it we had to be careful not to use properties in the proof of the converse that were not known. If we draw the circumcircle of EFGH it might be tempting to consider using properties of equal at the start, but that is not known (since we don’t yet know that I is the center of the circle). At the end, however, it is revealed that the circumcircle to EFGH is identical to the incircle in ABCD, with center I. Theorem 5.2. The incircles in the four overlapping triangles formed by the diagonals in a convex quadrilateral are tangent to the sides in eight points. The between such tangency points on each side of the quadrilateral are the vertices of a cyclic quadrilateral if and only if the original quadrilat- eral is tangential.

Figure 16. ABCD is tangential ⇔ M1M2M3M4 is cyclic

Proof. We use notations as in Figure 16, where S′, T ′, U ′, V ′, W ′, X′, Y ′, ′ Z are the tangency points and M1, M2, M3, M4 are the midpoints between tangency points. (⇒) In a tangential quadrilateral, S′Z′ = X′Y ′ according to Theorem ′ ′ 2.1 in [15], so M1Z = M4Y , and thus AM1 = AM4 according to the two ∠ 1 ∠ tangent theorem. This means that AM1M4 = 2 (π − A), so we get π − ∠A π − ∠B ∠A + ∠B ∠M2M1M4 = π − − = 2 2 2 34 Martin Josefsson and Mario Dalc´ın and in the same way ∠C + ∠D ∠M2M3M4 = . 2 Hence ∠M2M1M4 + ∠M2M3M4 = π, implying that M1M2M3M4 is cyclic. (⇐) We do an indirect proof of the converse. If ABCD is not a tangential quadrilateral, then we can assume without loss of generality that Z′S′ > ′ ′ ∠ 1 ∠ ∠ 1 ∠ X Y , so AM1M4 < 2 (π − A) and BM1M2 < 2 (π − B) since a longer side in a triangle is opposite a larger angle. In the same way ∠M2M3C < 1 ∠ ∠ 1 ∠ 2 (π − C) and M4M3D< 2 (π − D) since ′ ′ ′ ′ ′ ′ ′ ′ V W = Z S >X Y = T U where the first equality was proved in [15, p. 54] and the second is true by symmetry; they are both true in all convex quadrilaterals. Thus

∠M2M1M4 = π − ∠AM1M4 − ∠BM1M2 π − ∠A π − ∠B ∠A + ∠B >π − − = 2 2 2 and in the same way, we get ∠C + ∠D ∠M2M3M4 > . 2 Hence ∠A + ∠B + ∠C + ∠D ∠M2M1M4 + ∠M2M3M4 > = π 2 proving that M1M2M3M4 is not cyclic. 

In Figure 16 it appears as if the circumcircle to M1M2M3M4 is identical to the incircle in ABCD, but that is not true in general. The same holds for the circumcircle to M5M6M7M8 in Figure 17, where we have an excircle version of the previous theorem. Theorem 5.3. The four excircles to the four overlapping triangles formed by the diagonals in a convex quadrilateral that are tangent to the diagonals, are tangent to the extensions of the sides in eight points. The midpoints between such tangency points on each side of the quadrilateral are the vertices of a cyclic quadrilateral if and only if the original quadrilateral is tangential. Proof. We use notations as in Figure 17, where S′′, T ′′, U ′′, V ′′, W ′′, X′′, ′′ ′′ Y , Z are the tangency points and M5, M6, M7, M8 are the midpoints between tangency points on each side. When ABCD is tangential, then ′′ ′′ ′′ ′′ ′′ ′′ S Z = X Y according to Theorem 3.2 in [15]. Thus S M5 = X M8. The excircles to triangles ABC and ADC are tangent to AC at the same point, say Q, according to Theorem 3.1 in [15], so S′′A = QA = X′′A and ∠ 1 ∠ it follows that AM5 = AM8. Then AM5M8 = 2 (π − A) and in the same ∠ 1 ∠ way BM5M6 = 2 (π − B). We get π − ∠A π − ∠B ∠A + ∠B ∠M6M5M8 = π − − = 2 2 2 and in the same way, it holds ∠C + ∠D ∠M6M7M8 = . 2 More new characterizations of tangential quadrilaterals 35

Figure 17. ABCD is tangential ⇔ M5M6M7M8 is cyclic

Hence ∠M6M5M8 + ∠M6M7M8 = π, so M5M6M7M8 is cyclic. The converse result can be proved with an indirect proof in the same way as in the proof of the previous theorem. Writing down the details is left as an exercise for the reader. 

Next we have yet another characterization with a cyclic quadrilateral. Besides this one, there is at least one more property in Figure 18 that is easy to discover, which we will revisit in the final section. Can you spot it? Theorem 5.4. In a convex quadrilateral ABCD where the angle bisectors to A and C intersect at an interior point I, let O1, O2, O3, O4 be the circum- centers of triangles AIB, BIC, CID, DIA respectively. Then O1O2O3O4 is a cyclic quadrilateral if and only if ABCD is a tangential quadrilateral. Proof. Since the circumcenter to a triangle lies at the intersection of the perpendicular bisectors to the sides, we have O1O4 ⊥ AI and similar for the other distances BI, CI, DI (see Figure 18). This means that ∠O4O1O2 = π − ∠AIB and ∠O4O3O2 = π − ∠CID, so

∠O4O1O2 + ∠O4O3O2 = 2π − (∠AIB + ∠CID) holds in all convex quadrilaterals. According to Theorem 1 in [9], ABCD is tangential if and only if ∠AIB + ∠CID = π. Hence ABCD is tangential if and only if ∠O4O1O2 + ∠O4O3O2 = 2π − π = π confirming that ABCD is tangential if and only if O1O2O3O4 is cyclic.  36 Martin Josefsson and Mario Dalc´ın

Figure 18. ABCD is tangential ⇔ O1O2O3O4 is cyclic

6. Concurrent lines In this section we shall study five concurrences that are characterizations of tangential quadrilaterals. Theorem 6.1. Let the incircles in triangles ABC and ADC be tangent to the sides AB, BC, CD, DA of a convex quadrilateral ABCD at S′, T ′, W ′, X′ respectively. The lines S′T ′, X′W ′ and AC are concurrent if and only if ABCD is a tangential quadrilateral.

′ ′ ′ ′ Figure 19. ABCD is tangential ⇔ S T , X W , AC are concurrent

Proof. (⇒) In a tangential quadrilateral ABCD, let the line X′W ′ inter- sect the line AC at Q1. Applying Menelaus’s theorem (with non-directed distances) in triangle ACD with transversal X′W ′ yields ′ ′ AQ1 CW DX · ′ · ′ = 1 Q1C W D X A More new characterizations of tangential quadrilaterals 37 where DX′ = W ′D according to the two tangent theorem (see Figure 19). ′ ′ Likewise in triangle ACB, if the line S T intersects the line AC at Q2, then ′ ′ AQ2 CT BS · ′ · ′ = 1 Q2C T B S A where BS′ = T ′B. The two incircles are tangent to diagonal AC at the same point, say Q (Theorem 1 in [6]), so CW ′ = CQ = CT ′ and S′A = AQ = X′A. Hence ′ ′ AQ1 X A S A AQ2 = ′ = ′ = Q1C CW CT Q2C ′ ′ ′ ′ which proves that Q1 and Q2 are the same point, so S T , X W and AC are concurrent at Q1 = Q2. (⇐) In a convex quadrilateral that is not tangential, let Q′ and Q′′ be the tangency points of the two incircles on AC. Without loss of generality, we can assume that ′ ′ ′′ ′ CW = CQ < CQ = CT and ′ ′′ ′ ′ S A = AQ < AQ = X A; whence ′ ′ AQ1 X A S A AQ2 = ′ > ′ = Q1C CW CT Q2C proving that Q1 =6 Q2. 

Next we have a related characterization for the two incircles created by the other diagonal. This theorem was first discovered by the unknown referee to [10], and later independently rediscovered by the second author of the present paper. Theorem 6.2. Let the incircles in triangles ABD and BCD be tangent to the sides AB, BC, CD, DA of a convex quadrilateral ABCD at Z′, U ′, V ′, Y ′ respectively. The lines Z′U ′, Y ′V ′ and AC are concurrent if and only if ABCD is a tangential quadrilateral.

′ ′ ′ ′ Figure 20. ABCD is tangential ⇔ Z U , Y V , AC are concurrent 38 Martin Josefsson and Mario Dalc´ın

Proof. In a convex quadrilateral ABCD, let the line Y ′V ′ intersect the line ′ ′ AC at Q3 and the line Z U intersect the line AC at Q4 (see Figure 20). Applying Menelaus’s theorem in triangles ACD and ACB with transversals Y ′V ′ and Z′U ′ respectively, we get ′ ′ ′ ′ AQ3 CV DY AQ4 CU BZ (5) · ′ · ′ =1= · ′ · ′ . Q3C V D Y A Q4C U B Z A (⇒) In a tangential quadrilateral ABCD, the two incircles in triangles ABD and BCD are tangent to diagonal BD at the same point, say R (Theorem 1 in [6]). Thus V ′D = RD = DY ′ and U ′B = RB = BZ′. Together with the two tangent theorem, (5) is simplified as AQ AQ 3 = 4 Q3C Q4C ′ ′ ′ ′ which proves that Q3 and Q4 are the same point, so Z U , Y V and AC are concurrent at Q3 = Q4. (⇐) In a convex quadrilateral that is not tangential, let R′ and R′′ be the tangency points of the two incircles on BD. Without loss of generality, we have ′ ′′ ′ ′ V D = R D > R D = DY and ′ ′′ ′ ′ U B = R B < R B = BZ . Then, since CV ′ = CU ′ and Y ′A = Z′A, we get from (5) that ′ ′ ′ ′ AQ3 V D AQ3 DY AQ4 BZ AQ4 U B · ′ > · ′ = · ′ > · ′ Q3C V D Q3C V D Q4C U B Q4C U B where the inequality AQ AQ 3 > 4 Q3C Q4C implies that Q3 =6 Q4, completing the proof. 

Theorem 6.1 has the following excircle version. Theorem 6.3. Let the excircles to triangles ABC and ADC that are tan- gent to AC be tangent to the extensions of the sides AB, BC, CD, DA of a convex quadrilateral ABCD at S′′, T ′′, W ′′, X′′ respectively. The lines S′′T ′′, X′′W ′′ and AC are concurrent if and only if ABCD is a tangential quadrilateral. Proof. In a convex quadrilateral ABCD, let the line S′′T ′′ intersect the line ′′ ′′ AC at Q5 and the line W X intersect the line AC at Q6 (see Figure 21). Applying Menelaus’s theorem in triangles ABC and ADC with transversals S′′T ′′ and W ′′X′′ respectively, we get ′′ ′′ ′′ ′′ AQ5 CT BS AQ6 CW DX (6) · ′′ · ′′ =1= · ′′ · ′′ . Q5C T B S A Q6C W D X A (⇒) In a tangential quadrilateral ABCD, the two excircles to triangles ABC and ADC are tangent to diagonal AC at the same point, say M (Theorem 3.1 in [15]). Thus CT ′′ = CM = CW ′′ and S′′A = MA = X′′A. More new characterizations of tangential quadrilaterals 39

′′ ′′ ′′ ′′ Figure 21. ABCD is tangential ⇔ S T , X W , AC are concurrent

According to the two tangent theorem, T ′′B = BS′′ and W ′′D = DX′′, so (6) is simplified as AQ AQ 5 = 6 Q5C Q6C ′′ ′′ ′′ ′′ which proves that Q5 and Q6 are the same point. Hence S T , X W and AC are concurrent at Q5 = Q6. (⇐) In a convex quadrilateral that is not tangential, let M ′ and M ′′ be the tangency points of the two excircles on AC. Without loss of generality, we can assume that ′′ ′ ′′ ′′ CT = CM < CM = CW and ′′ ′ ′′ ′′ S A = M A>M A = X A.

Thus ′′ ′′ AQ5 S A X A AQ6 = ′′ > ′′ = Q5C CT CW Q6C proving that Q5 =6 Q6. 

Theorem 6.2 also has an excircle version. The proof of the following theorem is more or less identical to that of Theorem 6.2, so writing down the proof is left as an exercise for the reader. This excircle characterization is illustrated in Figure 22. Theorem 6.4. Let the excircles to triangles ABD and BCD that are tan- gent to BD be tangent to the sides AB, BC, CD, DA of a convex quadri- lateral ABCD at Z′′, U ′′, V ′′, Y ′′ respectively. The lines Z′′U ′′, Y ′′V ′′ and AC are concurrent if and only if ABCD is a tangential quadrilateral. The ‘if’ part of the next necessary and sufficient condition was Theorem 6 in [4], where it was proved in a different way. 40 Martin Josefsson and Mario Dalc´ın

′′ ′′ ′′ ′′ Figure 22. ABCD is tangential ⇔ Z U , Y V , AC are concurrent

Theorem 6.5. In a convex quadrilateral ABCD where the angle bisectors to B and D intersect at an interior point I, let E, F , G, H be the projections of I on the sides AB, BC, CD, DA respectively. The lines EF , GH, AC are concurrent if and only if ABCD is a tangential quadrilateral.

Proof. (⇒) The intersection I of two angle bisectors in a tangential quadri- lateral is the incenter, and E, F , G, H are the points where the incircle is tangent to the sides. Suppose that the extensions of GH and EF intersect the extension of diagonal AC in Q7 and Q8 respectively. Then applying Menelaus’s theorem in triangles ACD and ACB with transversals HG and EF respectively yields

AQ CG DH AQ CF BE 7 · · =1= 8 · · . Q7C GD HA Q8C FB EA

Here DH = GD, BE = FB, HA = EA and CG = CF according to the two tangent theorem. Thus

AQ AQ 7 = 8 Q7C Q8C which means that Q7 and Q8 divide AC in the same ratio, so they are the same point. Hence EF , GH, AC are concurrent at Q7 = Q8. (⇐) In a convex quadrilateral where EF , GH, AC are concurrent at a point Q, Menelaus’s theorem yields

AQ CG DH AQ CF BE · · =1= · · . QC GD HA QC FB EA More new characterizations of tangential quadrilaterals 41

Figure 23. If EF , HG, AC are concurrent, then ABCD is tangential

Triangles DHI and DGI are congruent (AAS), as are BEI and BFI, so this equality is simplified as HA · CF (7) CG = . EA Now let EI = FI = x and GI = HI = y (these distances are equal in pairs due to congruent triangles (AAS), see Figure 23). Applying the Pythagorean theorem in right triangles AEI and AHI, and also in CFI and CGI, we have EA2 + x2 = HA2 + y2, CF 2 + x2 = CG2 + y2. Thus HA2 − EA2 = x2 − y2 = CG2 − CF 2. Inserting (7), we get after clearing the denominator that EA2(HA2 − EA2)= CF 2(HA2 − EA2) which is factorized as (HA + EA)(HA − EA)(EA + CF )(EA − CF ) = 0. Then either HA = EA or EA = CF . In the first case it follows that HI = EI since triangles AHI and AEI are congruent (RHS). Thus FI = EI = HI = GI, so I is equidistant from the sides. Hence ABCD is tangential. In the second case, since also EB = FB, we have AB = CB. From (7), it follows that CG = HA, yielding AD = CD. Hence ABCD is a , which is always tangential. This completes the proof of the converse. 

7. Collinear points

The reader might have noticed in Figure 18 that it appears as if O1, I and O3 are collinear. That is true, and this is in fact another characterization of tangential quadrilaterals. 42 Martin Josefsson and Mario Dalc´ın

Theorem 7.1. In a convex quadrilateral ABCD where the angle bisectors to A and C intersect at an interior point I, let O1, O2, O3, O4 be the circumcenters of triangles AIB, BIC, CID, DIA respectively. Then O1, I and O3 are collinear if and only if ABCD is a tangential quadrilateral.

Figure 24. If ABCD is tangential, then O1, I, O3 are collinear

Proof. (⇒) When ABCD is a tangential quadrilateral, we have ∠BO1I = 2∠BAI = ∠A according to the inscribed angle theorem (see Figure 24). ∠ π ∠A ∠ π ∠D Thus O1IB = 2 − 2 . In the same way it holds that O3IC = 2 − 2 . ∠ π ∠B If F is the orthogonal projection of I on BC, then BIF = 2 − 2 and ∠ π ∠C FIC = 2 − 2 . Hence ∠O1IO3 = ∠O1IB + ∠BIF + ∠FIC + ∠O3IC ∠A + ∠B + ∠C + ∠D = 2π − 2 = π confirming that O1, I and O3 are collinear. (⇐) We prove the converse with an indirect proof. In a convex quadri- lateral ABCD that is not tangential, let E and F be the orthogonal pro- jections of I on AB and BC respectively (see Figure 25). Since IE =6 IF when ABCD is not tangential, we may assume without loss of generality that IE > IF . Next we draw a with BI that goes through F . Let E′ be a point on the semicircle such that triangles BEI and BE′I are congruent. Then we have that ∠EBI = ∠E′BI > ∠FBI since a longer chord in a circle is opposite a larger inscribed angle (IE′ > IF ). ∠ ∠B ∠ π ∠B ∠ ∠D Thus FBI < 2 , so BIF > 2 − 2 . In the same way CDI < 2 , so ∠ π ∠D ∠ π ∠A ∠ π ∠C O3IC > 2 − 2 , but we still have O1IB = 2 − 2 and FIC = 2 − 2 due to the angle bisectors at A and C. Hence ∠A + ∠B + ∠C + ∠D ∠O1IO3 > 2π − = π 2 More new characterizations of tangential quadrilaterals 43

Figure 25. If ABCD is not tangential, then O1, I, O3 are not collinear

so the points O1, I and O3 are not collinear. ∠ π There is another case, when EBI > 2 . Then point E is on the extension of AB and we don’t have to construct E′. It still remains true that ∠EBI > ∠FBI since ∠B<π in a convex quadrilateral. 

The similar result that O2, I and O4 are collinear if and only if ABCD is a tangential quadrilateral is of course also true. Then we have: Theorem 7.2. In a convex quadrilateral ABCD where the angle bisectors to A and C intersect at an interior point I, let O1, O2, O3, O4 be the circumcenters of triangles AIB, BIC, CID, DIA respectively. Then the diagonals in quadrilateral O1O2O3O4 intersect at I if and only if ABCD is a tangential quadrilateral. Proof. From the previous theorem and the similar result for the points O2, I and O4, we have that the diagonals in O1O2O3O4 intersect at I in a tangential quadrilateral (see Figure 24), and conversely: If ABCD is not a tangential quadrilateral, then neither of the diagonals of O1O2O3O4 goes through I. 

The line defined by the midpoints of the diagonals in a convex quadri- lateral is called Newton’s line. It is quite well known that the incenter of a tangential quadrilateral also lies on that line. We will formulate and prove a related characterization of tangential quadrilaterals concerning the intersec- tion point of two opposite angle bisectors in a convex quadrilateral. Before stating the theorem, let us remind the reader that a point P lies on Newton’s line in a convex quadrilateral ABCD if and only if

TAPB + TCPD = TBPC + TDPA according to Leon-Anne’s theorem, where TXYZ denote the area of triangle XYZ. One proof of this characterization of Newton’s line was given in [12, 44 Martin Josefsson and Mario Dalc´ın pp. 146–147]. In [17, pp. 134–135], it was proved that a convex quadrilateral is tangential if and only if TAIB + TCID = TBIC + TDIA where I is the point where two opposite internal angle bisectors intersect. That proof contained a minor flaw, which we will correct when using a large part of that argument to prove a related characterization: Theorem 7.3. If two opposite angle bisectors in a convex quadrilateral in- tersect at an interior point, then this point lies on Newton’s line if and only if it is a tangential quadrilateral. Proof. (⇒) In a tangential quadrilateral ABCD, a direct consequence of Pitot’s theorem is the equality AB · r CD · r BC · r DA · r + = + 2 2 2 2 where r is the inradius (the distance between the intersection I of two op- posite internal angle bisectors and the sides). Thus Leon-Anne’s theorem is satisfied, so I lies on Newton’s line.

Figure 26. ABCD is tangential ⇔ I ∈ MN

(⇐) In a convex quadrilateral where the intersection I of two opposite internal angle bisectors lies on Newton’s line MN, applying Leon-Anne’s theorem yields AB · x CD · y BC · y DA · x (8) + = + 2 2 2 2 where x is the distance between I and the sides AB and DA (see Figure 26), and similarly y is the distance between I and the sides BC and CD. Fac- toring (8), we have (9) x(AB − DA)= y(BC − CD). If AB = DA, then we get that BC = CD, so ABCD is a kite, which is tangential. Otherwise, assume without loss of generality that AB > DA. Then (9) implies that BC > CD. Now construct points E and F on AB and BC respectively such that AE = AD and CF = CD. Then triangles More new characterizations of tangential quadrilaterals 45

AEI and ADI are congruent (SAS), so EI = DI, and triangles CDI and CFI are also congruent, so DI = FI. Thus EI = FI. Next we rewrite (8) as AE · x BE · x CD · y BF · y CF · y DA · x + + = + + 2 2 2 2 2 2 which is simplified into BE · x BF · y (10) = 2 2 by the two pairs of congruent triangles. This area equality can also, by a well-known trigonometric area theorem, be expressed as BI · EI sin ∠BIE BI · FI sin ∠BIF = , 2 2 from which it follows that sin ∠BIE = sin ∠BIF . This equation has two possible solutions, either ∠BIE = ∠BIF or ∠BIE = π − ∠BIF . Let us consider the second case first (this was neglected in [17, p. 135]). We have that ∠D = ∠ADI +∠CDI = ∠AEI +∠CFI = ∠BIE +∠IBE +∠BIF +∠IBF where we applied the exterior angle theorem twice in the last step. Inserting ∠BIE = π − ∠BIF yields ∠D = π + ∠B, which is impossible in a convex quadrilateral (since any vertex angle is less than π), so this solution must be discarded. Hence the first case must be true. Then triangles BIE and BIF are congruent (SAS), so BE = BF . From (10), we get x = y, and then (9) implies AB − DA = BC − CD. This is just a reformulation of Pitot’s theorem, so we conclude that ABCD is a tangential quadrilateral. 

8. Concluding remarks As noted in the introduction, there are now 76 published necessary and sufficient conditions for when a convex quadrilateral can have an incircle that we know of. With such a huge number of characterizations, it is easy to get the impression that all properties of tangential quadrilaterals are characterizations. But that is not true. Let us consider one deviant property where the converse is not valid. In [3] it was proved that in any tangential quadrilateral ABCD, it holds that AC ⊥ MN where M is the intersection of the line BI and a line perpen- dicular to AB, N is the intersection of the line DI and a line perpendicular to AD, and I is the intersection of the angle bisectors to vertex angles B and D (see Figure 27). We have formulated the theorem a little differently from in [3] so that it would be possible to prove a converse by using the intersection of two opposite angle bisectors, which all convex quadrilaterals 46 Martin Josefsson and Mario Dalc´ın

Figure 27. AC ⊥ MN is not a sufficient condition for ABCD to be tangential have except the tilted kites (see Theorem 2.1 in [11]), instead of the in- center, which only tangential quadrilaterals have (as Grinberg did). Using equations derived in [3, pp. 1–2], we get MC2 − AM 2 − NC2 + AN 2 =(BC − AB)2 − (CD − DA)2 =(BC − AB + CD − DA)(BC − AB − CD + DA). It is a well-known characterization of orthodiagonal quadrilaterals (see Theo- rem 1 in [7]) that AC ⊥ MN if and only if MC2+AN 2 = AM 2+NC2, which is true if and only if either AB +DA = BC +CD or AB +CD = BC +DA. The second equality is Pitot’s characterization of tangential quadrilaterals, but the first equality is a characterization of one of the two types of extan- gential quadrilaterals (these have an exterior circle tangent to the extensions of all four sides, see [8]). Hence the converse statement that AC ⊥ MN im- plies ABCD to be a tangential quadrilateral is not true. We can conclude that many (in fact most) of the properties of tangential quadrilaterals are also characterizations of these quadrilaterals, but not all of them.

References

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SECONDARY SCHOOL KCM MARKARYD, SWEDEN E-mail address: [email protected]

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