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Croatian Mathematical Competitions

Geometry Difficulty 6.6 National Olympiad Prove it Croatia

Let $N$ be the foot of the altitude from $C$ to the hypotenuse $\overline{AB}$ of a right triangle $ABC$ . The angle bisectors of the angles $\angle NCA$ and $\angle BCN$ intersect the segment $\overline{AB}$ in the points $K$ and $L$ respectively. If $S$ and $T$ are incircles of the triangles $BCN$ and $NCA$ respectively, prove that the quadrilateral $KLST$ is cyclic. (from an article of A. Marić)

Solution

Denote $\triangle CAB = \alpha$ and $\triangle ABC = \beta$ . We have $\triangle BCN = \alpha$ and $\triangle ACN = \beta$ .

The lines $CS$ and $CT$ are angle bisectors of the angles $\triangle BCN$ and $\triangle ACN$ , so the points $C, S, L$ and $C, T, K$ are collinear. We have $\triangle BCK = \triangle BCN + \triangle NCK = \alpha + \frac{1}{2}\beta$ . Also, $\triangle BKC = \triangle CAK + \triangle ACK = \alpha + \frac{1}{2}\beta$ .

Hence $\triangle BCK = \triangle BKC$ , so the triangle $BCK$ is isosceles and we have $|BK| = |BC|$ . This implies that the point $S$ lies on the bisector of the segment $\overline{CK}$ , so $|SC| = |SK|$ . From this we conclude $\triangle SKT = \triangle SCT$ .

Analogously, the triangle $ACL$ is isosceles and the point $T$ lies on the bisector of the segment $\overline{CL}$ , so $\triangle SLT = \triangle SCT$ .

We have shown $\triangle SKT = \triangle SLT$ which means that the points $K, L, S$ and $T$ are concyclic.

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