Competition problem — Geometry — Maths Olympiad Prep

Library / /2 of 15

Geometry Difficulty 4.9 AIME Prove it Ukraine

On the sides $AB$ and $BC$ of the square $ABCD$ we marked points $K$ and $L$ correspondingly so that $KB = LC$ . Segments $AL$ and $CK$ intersect at point $P$ . Prove that lines $DP$ and $KL$ are perpendicular.

Solution

By construction $\Delta ADK = \Delta BAL$ . Then $\angle BAL + \angle ALB = \angle BAL + \angle AKD = 90^\circ$ , and $\Delta AMK$ has a right angle $M$ , so $AL \perp DK$ (fig. 14).

Similarly $DL \perp CK$ . Then lines $AL$ and $CK$ , which intersect at point $P$ , contain two altitudes of the $\Delta DKL$ , and therefore the line $DP$ has to contain the third altitude, so $KL \perp DP$ , as desired.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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