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Junior Turkish Mathematical Olympiad

Geometry Difficulty 3.5 AMC 10/12 Prove it Turkey

Let $ABCD$ be a cyclic quadrilateral and let the incenters of the triangles $BAD$ and $CAD$ be $I$ and $J$ , respectively. Let the intersection point of the line that passes through $I$ and perpendicular to $BD$ and the line that passes through $J$ and perpendicular to $AC$ be $K$ . Prove that $KI = KJ$ .

Solution

Let $E$ be the midpoint of arc $AD$ not containing points $B, C$ . Then, it is well known that $EA = ED = EI = EJ$ hence $A, I, J, D$ are concyclic. Therefore we get
$\angle JIK = \angle DIK - \angle DII = \angle DIK - \angle DAJ = 90^\circ - \frac{\angle ADB}{2} - \frac{\angle DAC}{2}$
Similarly $\angle IJK$ is equal to the same value and hence $KI = KJ$ .

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