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China Hong Kong Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Hong Kong

Let $D$ be a point on the side $BC$ of triangle $ABC$ such that $AB + BD = AC + CD$ . The line segment $AD$ cuts the incircle of triangle $ABC$ at $X$ and $Y$ with $X$ closer to $A$ . Let $E$ be the point of contact of the incircle of triangle $ABC$ on the side $BC$ . Show that
(i) $EY$ is perpendicular to $AD$ ,
(ii) $XD$ is $2IA'$ , where $I$ is the incentre of the triangle $ABC$ and $A'$ is the midpoint of $BC$ .

Solution

(i) Note that $AB + BD$ is the semiperimeter of $\triangle ABC$ , and so $D$ is the contact point of the $A$ -excircle of $\triangle ABC$ and $BC$ . Since $A$ is a centre of homothety between the incircle and the $A$ -excircle, $X$ and $D$ are corresponding points under this homothety. Therefore, the tangent at $X$ to the incircle is parallel to $BC$ . This implies $XE$ is a diameter of the incircle. It follows that $\angle XYE = 90^\circ$ , and hence $EY \perp AD$ .

(ii) Note that $I$ and $A'$ are midpoints of $EX$ and $ED$ respectively. By the midpoint theorem, $XD = 2IA'$ .

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