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Indian National Mathematical Olympiad

Geometry Difficulty 4.7 AIME Prove it India

Problem:

The in-circle of triangle $ABC$ touches the sides $BC$ , $CA$ and $AB$ in $K$ , $L$ and $M$ respectively. The line through $A$ and parallel to $LK$ meets $MK$ in $P$ and the line through $A$ and parallel to $MK$ meets $LK$ in $Q$ . Show that the line $PQ$ bisects the sides $AB$ and $AC$ of triangle $ABC$ .

Solution

Solution:

Let $AP$ , $AQ$ produced meet $BC$ in $D$ , $E$ respectively.

Since $MK$ is parallel to $AE$ , we have $\angle AEK = \angle MKB$ . Since $BK = BM$ , both being tangents to the circle from $B$ , $\angle MKB = \angle BMK$ . This with the fact that $MK$ is parallel to $AE$ gives us $\angle AEK = \angle MAE$ . This shows that $MAEK$ is an isosceles trapezoid. We conclude that $MA = KE$ . Similarly, we can prove that $AL = DK$ . But $AM = AL$ . We get that $DK = KE$ . Since $KP$ is parallel to $AE$ , we get $DP = PA$ and similarly $EQ = QA$ . This implies that $PQ$ is parallel to $DE$ and hence bisects $AB$ , $AC$ when produced.

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