IRL ABooklet

Geometry Difficulty 4.4 AIME Prove it Ireland

Consider three points $A$ , $B$ , $C$ on a circle $\Gamma$ , with $\angle BAC > 90^\circ$ . Let $d$ denote the line tangent to $\Gamma$ at $A$ . Points $M$ and $N$ are chosen on $d$ such that $\angle MBA = \angle ABC$ and $\angle NCA = \angle ACB$ .
Prove that $A$ is the midpoint of segment $MN$ .

Solution

Let $U$ be the intersection point of $BM$ and $CN$ . Then $A$ is the incentre of triangle $UBC$ , because $AB$ and $AC$ are angle bisectors by definition of $M$ and $N$ . In particular, $AU$ is the angle bisector of $\angle MUN$ .

Because $MN$ is tangent to $\Gamma$ , the Alternate Segment Theorem gives us
$\angle MAB = \angle ACB = \angle ACN$
$\angle NAC = \angle ABC = \angle ABM.$
Hence also $\angle BMA = \angle BAC = \angle ANC$ , which shows that $\triangle MUN$ is isosceles with $|MU| = |NU|$ . The angle bisector $AU$ is therefore a median as well, hence $A$ is the midpoint of $MN$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.