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Properties of Tilted Kites INTERNATIONAL JOURNAL OF GEOMETRY 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 quadrilaterals with two opposite equal angles. 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 kite is usually de…ned to be a quadrilateral with two distinct pairs of adjacent equal sides. It has several well-known properties, including: Two opposite equal angles; Perpendicular diagonals; A diagonal bisecting the other diagonal; An incircle (a circle tangent 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 tangential quadrilateral 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 ‘angle 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 vertex angle at D intersect the side \D BC at a point 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 trapezoid. Propertiesoftiltedkites 89 Figure 3: \A = \C i¤ EF GH is an isosceles trapezoid Proof. It is well known that the angle bisectors of the four angles in a convex quadrilateral either form a cyclic quadrilateral or they are concurrent. The latter is true only in a tangential quadrilateral, hence the exception for kites (including rhombi and squares). 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 rectangle or a square if the original quadrilateral is a parallelogram 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 trapezoids that are tilted kites are the parallelograms. 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 triangles 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 line 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 = \DP S and \Q = \BQS . The area 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 triangle 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 = \P SI2. \P Then from the exterior angle theorem, = v + and 2 = w + , so in 92 MartinJosefsson \P triangle P SI2, 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 DP T , we also have TB QB TD PD = ; = sin Q sin T sin P sin T where \T = \BT Q = \DT P . 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.
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