Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Soviet Union

Problem:

$ABCD$ is a parallelogram. The excircle of $ABC$ opposite $A$ has center $E$ and touches the line $AB$ at $X$ . The excircle of $ADC$ opposite $A$ has center $F$ and touches the line $AD$ at $Y$ . The line $FC$ meets the line $AB$ at $W$ , and the line $EC$ meets the line $AD$ at $Z$ . Show that $WX = YZ$ .

Solution

Solution:

We have the familiar result that $AY$ is perimeter $ADC$ (chase round using the fact that the two tangents from the same point have the same length). Similarly, $AX =$ perimeter $ABC =$ perimeter $ADC$ . So $AX = AY$ $(*)$

$AE$ is parallel to the bisector of $ACD$ , which is perpendicular to $CF$ . So $CW$ is perpendicular to $AE$ . Hence $AW = AC$ . Similarly $AZ = AC$ . Hence $AW = AZ$ . Subtracting from $(*)$ gives result.

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