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INTERNATIONAL JOURNAL OF Vol. 7 (2018), No. 1, 87 - 104

PROPERTIES OF TILTED KITES

MARTIN JOSEFSSON

Abstract. We prove …ve characterizations and derive several metric rela- tions in convex with two opposite equal . An interesting conclusion is that it is possible to express all quantities in such quadrilaterals in terms of their four sides alone.

1. Introduction A is usually de…ned to be a with two distinct pairs of adjacent equal sides. It has several well-known properties, including: Two opposite equal angles;  ;  A bisecting the other diagonal;  An incircle (a to all four sides);  An excircle (a circle tangent to the extensions of all four sides).  What quadrilaterals can we get by generalizing the kite? If we only con- sider one of the properties in the list at a time, then from the second we get an orthodiagonal quadrilateral (see [8]), from the third we get a bisect- diagonal quadrilateral (see [13]), and the fourth and …fth yields a and an extangential quadrilateral respectively (see [6] and [12]). But what about the …rst one? It seems perhaps to be the most basic. What properties hold in a quadrilateral that is de…ned only in terms of having two opposite equal angles? And what is it called? This class of quadrilaterals has been studied very scarcely. It is di¢ - cult to …nd any textbook or paper containing just one theorem or problem concerning such quadrilaterals, but they have been named at least twice in connection with quadrilateral classi…cations. Günter Graumann called them tilted kites in [3] and Michael de Villiers called them ‘ quadrilaterals’ in [18]. In this paper we use the former name and we shall study some fundamental properties of convex tilted kites (even though this restriction is not stated in every theorem). ————————————– Keywords and phrases: Tilted kite, Generalization, Characterization, Metric relations, Right quadrilateral, Kite (2010)Mathematics Subject Classi…cation: 51M04, 51M25 Received: 15.01.2018. In revised form: 23.03.2018. Accepted: 18.03.2018. 88 MartinJosefsson

Figure 1: A tilted kite with \A = \C

2. Characterizations We start with three simple necessary and su¢ cient conditions for a con- vex quadrilateral to be a tilted kite expressed in terms of di¤erent angle properties. Theorem 2.1. Two opposite angles in a convex quadrilateral are equal if and only if the angle bisectors of the other two angles are parallel.

Figure 2: \A = \C i¤ the angle bisectors to B and D are parallel

Proof. Assume the angle bisector to the angle at D intersect the side \D BC at a E. Then \CED =  2 \C (see Figure 2). The angle bisectors are parallel if and only if corresponding angles are equal, that is B D \ =  \ C B + D = A + B + C + D 2 C 2 2 \ , \ \ \ \ \ \ \ where we used the angle sum of a quadrilateral. Simplifying the last equality, we have that the angle bisectors are parallel if and only if the other two angles are equal (\A = \C).  Theorem 2.2. A convex quadrilateral, which is not a kite, is a tilted kite if and only if the angle bisectors of the four angles form an isosceles . Propertiesoftiltedkites 89

Figure 3: \A = \C i¤ EF GH is an

Proof. It is well known that the angle bisectors of the four angles in a convex quadrilateral either form a or they are concurrent. The latter is true only in a tangential quadrilateral, hence the exception for kites (including rhombi and ). Combining this with Theorem 2.1, we have that a convex quadrilateral is a tilted kite if and only if the angle bisectors of the four angles form a cyclic trapezoid, which is equivalent to an isosceles trapezoid (see Figure 3).  It can be noted that the formed isosceles trapezoid is a or a if the original quadrilateral is a or a rectangle respec- tively [16, p.34]. Theorem 2.3. Two opposite angles in a convex quadrilateral are equal if and only if the two angles between the extensions of opposite sides are equal.

Figure 4: A = C =  \ \ ,

Proof. Let the angles between the extensions of opposite sides in quadri- lateral ABCD be  and . If they intersect outside of D, then we have A + B + =  and B + C +  =  (see Figure 4). Hence A C = \ \ \ \ \ \ 90 MartinJosefsson

 , so A = C if and only if  = . In the same way we prove that \ \ \B = \D if and only if  = when the extended sides intersect outside of vertex A or C. A fourth case is when the intersection is outside of B, but that yields the same solution as the …rst. 

A limiting special case of this third characterization is that if one pair of opposite sides are parallel, then the quadrilateral is a tilted kite if and only if both pairs of opposite sides are parallel. That is to say that the only that are tilted kites are the . Next we have a more advanced characterization involving a tangent to two subtriangle incircles. The proof rests upon a remarkable formula that was posed as a problem by user ‘Fang-jh’at [2], but we do not know if this is the original discoverer. A few di¤erent proofs of this formula has been given at the website Art of Problem Solving; the idea behind the proof we give is due to the user ‘yetti’[19]. Theorem 2.4. In a convex quadrilateral ABCD, let L be the internal tan- gent other than AC to the incircles of ABC and ACD. Then L bisects diagonal BD if and only if the vertex angles at B and D are equal. Proof. Let L intersect BD at T . We shall derive the formula BT tan D (1) = 2 : DT B tan 2 Since T bisects BD if and only if BT = DT , which according to (1) is equivalent to \B = \D, the proof is then …nished.

Figure 5: \B = \D i¤ the tangent L bisects BD

We use the following notations: let the incircles of triangles ABC and ACD have radii r1 and r2, incenter I1 and I2 respectively, and let L intersect CD, AB and AC at P , Q and S respectively (see Figure 5). We have sin B = r1 and sin D = r2 . By the law of sines, 2 BI1 2 DI2 QB BI PD DI = 1 ; = 2 : sin \BI1Q sin \I1QB sin \DI2P sin \I2PD Propertiesoftiltedkites 91

It follows that (2) D B Q P QB BI sin BI Q sin I PD r sin sin 2 + 2 sin = 1 \ 1 \ 2 = 1 2 2 PD DI  sin I QB  sin DI P r  B   Q   D P 2 \ 1 \ 2 2 sin 2 sin 2 sin 2 + 2 where we have used the notations \P = \DPS and \Q = \BQS . The of triangles AI1C and AI2C are respectively 1 1 AC r1 = AI1 CI1 sin AI1C; 2  2   \ 1 1 AC r2 = AI2 CI2 sin CI2A 2  2   \ and, according to the law of sines applied in triangles AI1I2 and CI1I2, we also have AI AI CI CI 1 = 2 ; 1 = 2 : sin \AI2S sin \SI1A sin \SI2C sin \CI1S Thus r AI CI sin AI C sin AI S sin SI C sin AI C (3) 1 = 1 1 \ 1 = \ 2 \ 2 \ 1 : r2 AI2  CI2  sin \CI2A sin \SI1A  sin \CI1S  sin \CI2A Using the sum of angles in triangles AI1C and ABC, we get \AI1C + \I1AC + \I1CA =  and 2\I1AC + 2\I1CA + \B = ; whence \AI1C =  B  D + \ . Similarly, CI2A = + \ . Since CSQ = QAS +  Q and 2 2 \ 2 2 \ \ \ the circle with center I1 is an excircle to AQS, it holds that \QAS \QAS  \Q  \Q SI1A =   + = : \ 2 2 2 2 2 2    P Similarly, SI2C = \ . It follows that \ 2 2  \B  \Q \B \Q CI1S = AI1C SI1A = + = + \ \ \ 2 2 2 2 2 2   \D \P and in the same way \AI2S = 2 + 2 .

Figure 6: Angle notations in quadrilateral CPI2S

Next, consider the quadrilateral CPI2S (see Figure 6). Let us denote v = \SI2C, w = \PI2C, = \PCI2 = \SCI2 and  = \ASI2 = \PSI2. \P Then from the exterior angle theorem,  = v + and 2 = w + , so in 92 MartinJosefsson

\P triangle PSI2, we have v + w + 2 +  = . Eliminating w and , we get v =  \P , so 2 2  P P sin SI2C = sin v = sin = cos : \ 2 2 2   Q In the same way sin \SI1A = cos 2 . Inserting the six angle expressions we have just derived into (3) yields D P P B r1 sin 2 + 2 cos 2 cos 2 = Q D ; r2 cos  B Q  cos 2  sin 2 + 2 2 which in turn imply that (2) can be simpli…ed as QB cos P cos B sin D sin P sin P tan D (4) = 2 2 2 2 = 2 PD Q  D  B  Q sin Q  B cos 2 cos 2 sin 2 sin 2 tan 2 1 x x where we used the double angle formula 2 sin x = sin 2 cos 2 in the last equality. According to the law of sines applied in triangles BQT and DPT , we also have TB QB TD PD = ; = sin Q sin T sin P sin T where \T = \BTQ = \DTP . We …nally get from these and (4) that TB QB sin Q tan D = = 2 TD PD sin P B  tan 2 which completes the proof. 

The following formula was derived in [15, p.194], wherein it is attributed to Ovidiu Pop. However, he did not consider the converse. This is the only metric formula in this paper that is independent on which pair of opposite angles in the quadrilateral that are equal. Theorem 2.5. A convex quadrilateral with consecutive sides a, b, c, d is a tilted kite if and only if the distance v between the of the diagonals is given by 1 (ab cd)(bc ad) v = : 2r ac bd This characterization does not apply to kites. Proof. Consider a convex quadrilateral ABCD where E and F are the midpoints of diagonals AC and BD respectively, and N is the of side AD. We have that EN and FN are midlines in triangles ADC 1 1 and DAB respectively. Thus EN = 2 c, FN = 2 a, \ANE = \D and \DNF = \A (see Figure 7). It holds that \ENF =  (\A + \D) . Using the law of cosines in triangle EFN yields j j a2 c2 a c v2 = + 2 cos  (A + D) ; 4 4  2  2 j j whence 4v2 = a2 + c2 + 2ac cos (A + D). In the same way (using triangle EFK, where K is the midpoint of AB) it is obtained that 4v2 = b2 + d2 + Propertiesoftiltedkites 93

Figure 7: The distance between the diagonal midpoints

2bd cos (A + B). Thus in a tilted kite where \B = \D we get by combining the last two formulas that 4v2(ac bd) = ac(b2 + d2) bd(a2 + c2) and the formula in the theorem follows after factorization. We get the same formula in the other case when \A = \C, since the only di¤erence is the change b d which gives two sign changes in the numerator that cancel. The exception$ for kites is explained later in Corollary 4.1. This is in the case when the denominator is zero. For the converse, we use that 4v2 = a2 + c2 + 2ac cos (A + D) and 4v2 = b2 + d2 + 2bd cos (A + B) hold in all convex quadrilaterals according to the …rst part of the proof. Thus we get 4v2(ac bd) = ac(b2 + d2) bd(a2 + c2) + 2abcd cos (A + B) cos (A + D) : But the assumption is that 4v2(ac bd) = ac(b2 + d2) bd(a2 + c2), so in the quadrilateral(s) we seek it must hold that 2abcd cos (A + B) cos (A + D) = 0 which according to one of the sum-to-product formulas is equivalent to 2A + B + D D B 4abcd sin sin = 0; 2 2 or using A + B + C + D = 2 and sin ( + ') = sin ', we get \ \ \ \ A C D B 4abcd sin sin = 0: 2 2 A C D B Now either sin 2 = 0 or sin 2 = 0. The …rst gives the solutions \A = \C or \A = 2 + \C, where the second is not valid in a convex quadrilateral. In the same way the second sine-equation only has one valid solution, \D = \B. This proves that the quadrilateral is a tilted kite. 

3. Circumcircle and incircle One special case of a tilted kite is a quadrilateral with two opposite right angles. It was called a right quadrilateral in [18, pp.73–74], a ‘Hjelmslev quadrilateral’ in [5, p.430] and a ‘right Pythagorean quadrilateral’ in [11, p.220]. We will use the …rst name in this paper (see Figure 8). 94 MartinJosefsson

Figure 8: A right quadrilateral is a cyclic tilted kite

Theorem 3.1. A tilted kite has a circumcircle if and only if it is a right quadrilateral. Proof. A convex quadrilateral has a circumcircle if and only if its opposite angles are supplementary angles. The system of equations \A+\C =  and  \A = \C has the solution \A = \C = 2 , that is, a right quadrilateral.  All kites have an incircle, but are there any other more general tilted kites with an incircle that are not kites? That question is answered with ‘no’by the next theorem. Theorem 3.2. A tilted kite has an incircle if and only if it is a kite. Proof. A convex quadrilateral has an incircle if and only if its internal angle bisectors are concurrent. We know from Theorem 2.1 that a pair of opposite angle bisectors in a tilted kite are parallel. For them to intersect, these angle bisectors must coincide. This means that they constitute a diagonal, and this diagonal divide the tilted kite into two congruent triangles (a characterization of kites). Hence a tilted kite has an incircle if and only if it is a kite.  4. Metric relations concerning distances We shall use the following notations in the rest of this paper: a tilted kite ABCD has the sides a = AB, b = BC, c = CD, d = DA and the diagonals p = AC, q = BD. There are two main types of tilted kites, with \A = \C or \B = \D. Most formulas are derived for and only valid in the …rst case. In the second case, similar formulas hold which can be derived in the same way, or we get them by when interchanging b and d in the formulas for the …rst case. The following metric relation is known to hold in a cyclic quadrilateral (see [17, p.1]), but it is equally true in all tilted kites. Theorem 4.1. In a convex quadrilateral ABCD where the diagonals inter- sect at P , we have AP DA AB =  CP BC CD  Propertiesoftiltedkites 95 if and only if the quadrilateral is either a tilted kite or a cyclic quadrilateral.

Figure 9: Heights in subtriangles ABD and BCD

Proof. Let T1 and T2 be the of triangles ABD and BCD respectively, and h1 and h2 their heights to the common side BD (see Figure 9). Then we have AP h1 T1 DA AB sin A = = =  : CP h2 T2 BC CD sin C Hence the equality in the theorem is true if and only if sin A = sin C, which is equivalent to \A = \C or \A =  \C. The …rst solution is the de…nition of a tilted kite and the second is a well-known characterization of cyclic quadrilaterals.  Next we derive formulas for the diagonal lengths in a tilted kite. Theorem 4.2. In a tilted kite ABCD with consecutive sides a, b, c, d where \A = \C, the diagonal opposite the equal angles has the length (ab cd)(ac bd) q = bc ad r and the diagonal connecting the vertices with equal angles has the length (a2 b2)2cd ab(c2 d2)2 p = : (ac bd)(ad bc) s Neither formula is valid in a kite nor in a parallelogram.

Figure 10: The diagonal q 96 MartinJosefsson

Proof. By the law of cosines we have q2 = a2 + d2 2ad cos A and q2 = b2 + c2 2bc cos A since A = C (see Figure 10). From these, we get \ \ q2(bc ad) = ab(ac bd) cd(ac bd) and the …rst formula follows directly. Using Euler’squadrilateral theorem (see [15, p.14])

(5) a2 + b2 + c2 + d2 = p2 + q2 + 4v2 in combination with Theorem 2.5 and the formula we just derived yields (ab cd)(bd ac) (ab cd)(ad bc) p2 = a2 + b2 + c2 + d2 ad bc bd ac abc4 + a4cd 2a2b2cd + b4cd + 2abc2d2 abd4 = (ac bd)(ad bc) from which the second formula follows by collecting similar terms. The cases of exception are proved in Corollaries 4.1 and 4.2.  Corollary 4.1. A tilted kite with consecutive sides a, b, c, d is a kite if and only if ac = bd.

Proof. ( ) If a tilted kite is a kite, then by de…nition a = b and c = d or a = d and)b = c. It follows that ac = bd. ( ) We prove the converse in the case when diagonal q is opposite the ( equal angles (\A = \C); the other case is similar. By Theorem 4.2, q2(bc ad) = (ac bd)(ab cd): Since q = 0, we get that ac = bd imply ad = bc. Solving these two simul- taneous6 equations yields a = b and c = d, which means the tilted kite is a kite.  Corollary 4.2. In a tilted kite with consecutive sides a, b, c, d, it holds that ab = cd or ad = bc if and only if it is either a kite or a parallelogram.

Proof. From Theorem 2.5 we have that

(ab cd)(bc ad) = 4v2(ac bd): Hence ab = cd or ad = bc is equivalent to v = 0 or ac = bd. The …rst equality is a well-known characterization for a parallelogram (bisecting diagonals) and the second is a necessary and su¢ cient condition for when a tilted kite is a kite according to the previous corollary. 

From now on, whenever there is a case of exception in a theorem regarding a kite or a parallelogram, then it is due to Corollary 4.1 or 4.2. A bimedian in a quadrilateral is a connecting the midpoints of opposite sides. The bimedian connecting sides a and c is labeled m, and the one connecting b and d is labeled n. Propertiesoftiltedkites 97

Theorem 4.3. In a tilted kite ABCD with consecutive sides a, b, c, d where \A = \C, the lengths of the bimedians are 1 (ab + cd)(ad + bc) 2bd(b2 + d2) m = ; 2 ac bd r 1 (ab + cd)(ad + bc) 2ac(a2 + c2) n = 2 bd ac r if it is not a kite.

Proof. In [7, p.162] it was noted that the lengths of the bimedians in a convex quadrilateral satisfy the formulas 4m2 = 2(b2 + d2) 4v2; 4n2 = 2(a2 + c2) 4v2: Inserting the expression for 4v2 from Theorem 2.5 and simplifying directly yields the formulas in this theorem. 

5. Area The area formula for a tilted kite is certainly not well known. According to [1, p.34], this formula appeared in a German collection of geometrical problems by Meier Hirsch in 1805, but not separated into two parts as we have done.

Theorem 5.1. A tilted kite ABCD with \A = \C and consecutive sides a, b, c, d has the area ad + bc K = Q ad bc j j if it is neither a kite nor a parallelogram, where

Q = 1 ( a + b + c d)(a b + c d)(a + b c d)(a + b + c + d) : 4 Proof. Ap tilted kite with \A = \C (see Figure 10) has the area 1 1 1 (6) K = 2 ad sin A + 2 bc sin C = 2 sin A(ad + bc): Applying the law of cosines in the two triangles created by diagonal BD yields a2 + d2 2ad cos A = b2 + c2 2bc cos C, that is a2 + d2 b2 c2 = 2 cos A(ad bc): From the trigonometric version of the , it holds 4K2 (a2 + d2 b2 c2)2 1 = sin2 A + cos2 A = + : (ad + bc)2 4(ad bc)2 Factoring this equality, we get that the expression 16K2(ad bc)2 (ad + bc)2 98 MartinJosefsson equals 4(ad bc)2 (a2 + d2 b2 c2)2 = (a + d)2 (b + c)2 (b c)2 (a d)2 = (a + d + b + c)(a + d b c)(b c + a d)(b c a + d)   = (a + d + b + c)( a d + b + c)(b c + a d)( b + c + a d) and Hirsch’s formula follows. It is not valid in kites and parallelograms according to Corollary 4.2.  It is noteworthy that Q has a geometrical meaning. It gives the area of a crossed cyclic quadrilateral with consecutive sides a, b, c, d according to [14, p.500]. The formula for Q is the same as Brahmagupta’s formula with consecutive sides a, b, c, d. It is easy to see, by factoring out 1 from two of the parentheses at a time, that it is possible to have any one of the four sides as the negative. Corollary 5.1. In a tilted kite with sides of di¤erent lengths, the sum of the longest and the shortest side must be less than the sum of the other two sides. Proof. There are several cases, but suppose …rst the consecutive sides a, b, c, d have this order of magnitude from longest to shortest. Then it is obvious that a b + c d > 0 and a + b c d > 0. In order for the radicand of Q to be positive, we must have a +b + c d > 0, from which the conclusion follows in this case. It is always possible to rename the sides so that a is the longest side. Then there are only six cases to consider. Reasoning as in the …rst case, we get the results indicated in Table 1, which concludes the proof. 

Side inequalities Conclusion a > b > c > d b + c > a + d a > b > d > c b + d > a + c a > c > b > d b + c > a + d a > c > d > b c + d > a + b a > d > b > c b + d > a + c a > d > c > b c + d > a + b

Table 1: The six di¤erent cases of side inequalities

Corollary 5.2. A tilted kite ABCD with \A = \C and consecutive sides a, b, c, d, that is neither a kite nor a parallelogram, has the area

ad + bc 1 a2 b2 c2 + d2 2 K = 1 : 2 s 4 ad bc   The formula is simpli…ed to ad + bc K = 2 Propertiesoftiltedkites 99

 if and only if the tilted kite is a right quadrilateral with \A = \C = 2 . Proof. The …rst formula is a direct consequence of the proof of Theorem 5.1. From a2 + d2 b2 c2 = 2 cos A(ad bc) we get that a2 + d2 = b2 + c2  if and only if \A = 2 (a right quadrilateral) since it is neither a kite nor a parallelogram, so ad = bc according to Corollary 4.2. 6 

6. Angle formulas There are several di¤erent formulas for calculating the vertex angles in a tilted kite. Theorem 6.1. The two equal angles in a tilted kite ABCD with consecutive sides a, b, c, d are given by a2 b2 c2 + d2 cos A = = cos C 2(ad bc) and the other two angles are given by 4abcd + (c2 d2)2 (a2 + b2)(c2 + d2) cos B = ; 2(ac bd)(ad bc) 4abcd + (a2 b2)2 (a2 + b2)(c2 + d2) cos D = : 2(ac bd)(bc ad) Neither formula is valid in a kite nor in a parallelogram. Proof. By the law of cosines and Theorem 4.2, a2 + b2 p2 4abcd + (c2 d2)2 (a2 + b2)(c2 + d2) cos B = = : 2ab 2(ac bd)(ad bc) The derivations of the other formulas are similar.  Theorem 6.2. The two equal angles in a tilted kite ABCD with consecutive sides a, b, c, d are given by

A (a + b c d)(a b + c d) C tan = = tan 2 ( a + b + c d)(a + b + c + d) 2 s and the other two angles are given by

B c + d ( a + b + c d)(a b + c d) tan = ; 2 c d (a + b c d)(a + b + c + d) j js D a + b ( a + b + c d)(a b + c d) tan = : 2 a b (a + b c d)(a + b + c + d) j js Proof. We derive the formula for angle B; the other proofs are similar. Since B 1 cos B (7) tan2 = 2 1 + cos B   100 MartinJosefsson we will …rst simplify expressions for the numerator and denominator in this formula. From Theorem 6.1, we have that 1 cos B equals 2(ac bd)(ad bc) 4abcd (c2 d2)2 + (a2 + b2)(c2 + d2) 2(ac bd)(ad bc) (a2 2ab + b2)(c2 + 2cd + d2) (c2 d2)2 = 2(ac bd)(ad bc) (c + d)2(a b + c d)(a b c + d) (8) = : 2(ac bd)(ad bc) In the same way we have that (c d)2(a + b + c + d)( a b + c + d) (9) 1 + cos B = : 2(ac bd)(ad bc) Inserting these expressions into (7) results after simpli…cation in the formula B we seek, since tan 2 > 0.  Theorem 6.3. The two congruent angles in a tilted kite ABCD with con- secutive sides a, b, c, d and area K are given by 2K sin A = = sin C ad + bc and the other two angles are given by 2K c2 d2 sin B = ; ad + bc ac bd 2K a2 b2 sin D = : ad + bc ac bd

The last two formulas are not valid in a kite.

Proof. The …rst two formulas is just a rewrite of (6). For the third formula, using (8) and (9), we get sin2 B = 1 cos2 B = (1 + cos B)(1 cos B) (c d)2(a + b + c + d)( a b + c + d) = 2(ac bd)(ad bc) (c + d)2(a b + c d)(a b c + d)  2(ac bd)(ad bc) (c2 d2)2 16Q2 =  4(ac bd)2(ad bc)2 where we used the de…nition of Q from Theorem 5.1 in the last equality. Whence c2 d2 2Q c2 d2 2K sin B = j j  = j j ac bd ad bc ac bd  ad + bc j jj j j j according to Theorem 5.1. The formula for sin D is derived in the same way.  There is also a formula for the measure of the angles between the ex- tensions of opposite sides. Remember that these two angles are equal by Theorem 2.3. Propertiesoftiltedkites 101

Theorem 6.4. The two equal angles  and between the extensions of opposite sides in a tilted kite with consecutive sides a, b, c, d are given by

 1 (a + b c d)( a + b + c d) sin = 2 2 ac bd r if it’s not a kite.

Proof. We will use Euler’s quadrilateral theorem (5) and Theorem 10 in [9], according to which p2 + q2 = b2 + d2 + 2ac cos , where  is the angle between the extensions of the sides a and c. Both of these equalities are valid in all convex quadrilaterals. Combining them yields 4v2 = a2 + b2 + c2 + d2 (b2 + d2 + 2ac cos ) = (a c)2 + 2ac(1 cos )  = (a c)2 + 4ac sin2 : 2   Now inserting Theorem 2.5 and solving for the angle, we get  (ab cd)(bc ad) (a c)2(ac bd) 4ac sin2 = 2 ac bd   ac(b d + a c)(b d a + c) = ac bd which simpli…es to the formula in the theorem. 

7. The area of the diagonal point triangle If the diagonals in a quadrilateral intersect at P and the extensions of the opposite sides intersect at E and F , then triangle EFP is called the diagonal point triangle (see Figure 11). In order for this triangle to exist, we must assume the quadrilateral has no pair of opposite parallel sides.

Figure 11: The diagonal point triangle EFP 102 MartinJosefsson

Theorem 7.1. In a tilted kite ABCD with \A = \C, which have the consecutive sides a, b, c, d and area K, the diagonal point triangle has the area 2abcd a2 b2 c2 d2 K T = (ad + bc)2(ad bc)2 if the tilted kite is neither a kite nor a parallelogram. Proof. According to Richard Guy’sversion of Hugh ApSimon’sformula (see [4, p.163]), the diagonal point triangle belonging to a convex quadrilateral ABCD has the area 2T T T T (10) T = 1 2 3 4 K(T1T2 T3T4) where T1, T2, T3, T4 are the areas of triangles ABC, ACD, ABD, BCD respectively. In a tilted kite with \A = \C, we have that ab abK c2 d2 T = sin B = ; 1 2 ad + bc ac bd cd cdK a2 b2 T = sin D = ; 2 2 ad + bc ac bd ad adK T = sin A = ; 3 2 ad + bc bc bcK T = sin C = 4 2 ad + bc where we used Theorem 6.3. Thus 2(abcd)2K4 a2 b2 c2 d2 2T T T T = 1 2 3 4 (ac bd)2(ad + bc)4 and abcdK3 a2 b2 c2 d2 (ac bd)2 K T1T2 T3T4 = 2 2 : j j (ad + bc) (ac bd)

We put an absolute value for T1T2 T3T4 to cover all cases. Now there are two possibilities. Depending on the sign of the inner absolute value in the numerator in the last expression, that numerator a2 b2 c2 d2 (ac bd)2 j jj j is either simpli…ed to (ad bc)2 or (ad + bc)2 2( a2c2 + b2d2) depend- ing on whether (a2 b2)(c2 d2) is positive or negative respectively. To decide the correct alternative, we study a rectangle. Note that since oppo- site sides in a rectangle are parallel, its diagonal point triangle must have in…nite area. Thus we expect the denominator in the formula for the area of the diagonal point triangle to be zero. Inserting a = c and b = d, we get that (ad bc)2 = 0, but the second possible expression is simpli…ed to 2(a2 b2)2 = 0. Whence the correct area formula is

6 2(abcd)2K4 a2 b2 c2 d2 (ad + bc)2(ac bd)2

T = 2 4 3 2 (ac bd ) (ad + bc)  abcdK (ad bc) which is simpli…ed to the formula in the theorem.  Propertiesoftiltedkites 103

8. Concluding remarks In [10, p.73] we noted that the two classes of generalized kites that are quadrilaterals where one diagonal is bisected by the other diagonal (later called a bisect-diagonal quadrilateral) and where two opposite angles are equal (a tilted kite) have appeared rarely in the literature on geometry. The former was then explored in [13] and the latter in the present paper. An important feature of these quadrilaterals is a property they share: that the metric relations for all important quantities such as the two diagonal lengths, the distance between the diagonal midpoints, and the quadrilateral area can be expressed in terms of the quadrilateral sides alone. Furthermore, their area formulas are surprisingly similar. The area of a bisect-diagonal quadri- lateral can according to the corollary to Theorem 10 in [13] be expressed as ad bc 2 K = S 1 4 s a2 b2 c2 + d2   when diagonal p is bisected by diagonal q. Here S stands for the area of a cyclic quadrilateral (Brahmagupta’s formula). The area of a tilted kite ABCD can according to the corollary to Theorem 5.1 in the present paper be expressed as

1 a2 b2 c2 + d2 2 K = S 1 s 4 ad bc   1 when \A = \C. Here S = 2 (ad + bc) is the area of a right quadrilateral (which is a cyclic tilted kite). We …nd the similarity between these formulas very interesting and remarkable.

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