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Serbian Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Serbia

Problem:

Let $D$ be a point on side $AC$ of triangle $ABC$ in which $AB < BC$ such that $AB = BD$ . The circle inscribed in $\triangle ABC$ touches $AB$ at $K$ and $AC$ at $L$ , and $J$ is the center of the inscribed circle of triangle $BCD$ . Prove that $KL$ bisects the segment $AJ$ .

Solution

Solution:

Let $M$ be a point on $AC$ such that $JM \parallel KL$ . It suffices to prove that $AM = 2AL$ .

From $\angle BDA = \alpha$ we obtain $\angle JDM = 90^\circ - \frac{\alpha}{2} = \angle KLA = \angle JMD$ , so $JM = JD$ , and the point of tangency of the inscribed circle of $\triangle BCD$ with $CD$ is the midpoint $T$ of segment $MD$ . Therefore, $DM = 2DT = BD + CD - BC = AB - BC + CD$ , from which
$AM = AD + DM = AC + AB - BC = 2AL$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from sr; metadata (topic, difficulty) added by this project.