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New Zealand Mathematical Olympiad

Geometry Difficulty 6.0 National Olympiad Prove it New Zealand

Problem:

Triangle $ABC$ is right-angled at $B$ and has incentre $I$ . Points $D$ , $E$ and $F$ are the points where the incircle of the triangle touches the sides $BC$ , $AC$ and $AB$ respectively. Lines $CI$ and $EF$ intersect at point $P$ . Lines $DP$ and $AB$ intersect at point $Q$ . Prove that $AQ = BF$ .

Solution

Solution:

First note that $ID = IE = IF$ because they are all radii of the incircle, and $\angle BFI = \angle BDI = 90^{\circ}$ because tangents are perpendicular to radii. Since $\angle ABC = 90^{\circ}$ we have $BFID$ a square and so $BD = BF = ID$ too. Thus $\triangle EIF$ is isosceles and so $\angle IFE = \angle FEI$ .

Since $CI$ is the bisector of $\angle DCE$ , we see that $D$ and $E$ are reflections of each other over line $CIP$ . Therefore $\angle IDP = \angle PEI$ . Hence
$\angle IDP = \angle FEI = \angle IFP$
and therefore quadrilateral $FPID$ is cyclic. Therefore $\angle FPD = \angle FID = 90^{\circ}$ ( $BFID$ is a square). Since $AI$ is the bisector of $\angle EAF$ , we see that $E$ and $F$ are reflections of each other over line $AI$ . Therefore $EF \perp AI$ . Hence
$AI \parallel DQ$
because they are both perpendicular to $EF$ . We also have $AQ \parallel ID$ (because $BFID$ is a square) so $QAID$ is a parallelogram. Therefore
$AQ = ID = BF$
as required.

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