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Iranian Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Iran

Let $ABC$ be an acute-angled triangle and $D$ be the foot of altitude from $A$ . Let $K$ , $L$ be the touching points of tangent lines from $D$ to the circles with diagonals $AB$ and $AC$ , respectively. Point $S$ is given on the plane such that
$\angle ABC + \angle ABS = \angle ACB + \angle ACS = 180^\circ.$
Prove that $A$ , $K$ , $L$ and $S$ lie on a circle.

Solution

Suppose that $L$ and $K$ lie on the circumcircle of $ABD$ and $ADC$ , respectively.

Note that $\angle LAB = 90^\circ - \angle ABL = 90^\circ - \angle ADL = 90^\circ - \angle ACB$ . Similarly we have $\angle CAK = 90^\circ - \angle CBA$ . Summing these two implies that $\angle LAK = 2\angle BAC$ . Denote by $T$ the intersection of $BL$ , $CK$ , then
$\angle TBC = 180^\circ - \angle CBA - \angle ABL = 180^\circ - \angle CBA - \angle ACB = \angle BAC,$
and similarly $\angle TCB = \angle BAC$ . Therefore $ABC$ is an isosceles triangle and $\angle CTB = 180^\circ - 2\angle BAC$ . Similarly, one can show that $\angle CSB = 180^\circ - 2\angle BAC$ and so quadrilateral $BCTS$ is cyclic. Properties of $S$ imply that $A$ is the $S$ -excenter of the triangle $BCS$ and so $SA$ is the angle bisector of $\angle BSC$ . Let $M$ be the midpoint of the arc $BC$ (the one that does not contain $S$ ) in the circumcircle of $BCS$ . Clearly $M$ lies on $SA$ .
On the other hand, since triangle $BTC$ is isosceles, $MT$ is a diagonal of the circumcircle of $BCS$ . Thus, we can conclude that
$\angle AST = \angle MST = 90^\circ = \angle AKT = \angle ALT.$
This implies $AKLS$ is cyclic as desired. ■

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