Let $ABCD$ be a rectangle. Two perpendicular lines pass through point $B$. One of them meets segment $AD$ at point $K$, and the second one meets the extension of side $CD$ at point $L$. Let $F$ be the common point of $KL$ and $AC$. Prove that $BF\perp KL$.
Problem
Source: Sharygin Geometry Olympiad 2014 - Problem 8
Tags: geometry, rectangle, Sharygin Geometry Olympiad
Pirkuliyev Rovsen
15.11.2014 20:13
See here http://geometry.ru/olimp/2014/zaochsol-e.pdf
djmathman
15.11.2014 20:14
^^ These are being posted for Contests page purposes.
I_m_vanu1996
19.11.2014 05:14
Clearly, $KBLD$ is cyclic,so $\angle {BKL}=\angle {BDL}$,but $\angle {BAC}=\angle {BDC}$,hence $\angle {BAF}=\angle {BKF}$,so $ABFK$ is also cyclic,but $\angle {BAK}=90$,so $\angle {BFK}=90$.
jayme
19.11.2014 10:04
Dear Mathlinkers, in order to avoid some angles... why not the pivot theorem applied to the triangle CLF and the menelian DKA... Sincerely Jean-Louis
PROF65
19.11.2014 14:10
You can use the rotation with center B and angle 90 and deduce A,B,F and K are cyclic which close the proof
parmenides51
01.08.2018 16:22
a coordinate solution given here
OlympusHero
25.06.2021 03:07
Already written in a different thread. Posting for storage.
We have $\angle D = 90^\circ$ and $\angle KBL = 90^\circ$, so $BLDK$ is cyclic. Hence $\angle BKL = \angle BDL = \angle BAF$, implying that $ABFK$ is cyclic. Hence $\angle A + \angle BFK = 180^\circ$, so $\angle BFK = 90^\circ$ or $BF \perp KL$ as desired.
lifeismathematics
19.04.2023 12:08
mark $\angle{BLC}=\theta$ , then we get $\angle{KBA}=90^{\circ}-\theta \implies \angle{BKA}=\theta$ so $\triangle{ABK}\sim \triangle{CBL}$ now from this we get $\frac{CB}{BA}=\frac{BL}{BK}$ which gives $\triangle{ABC} \sim \triangle{KBL}$, so we have $\angle{BLK}=\angle{BCA}$ giving quadrilateral $BLCF$ is cyclic and hence $\angle{BFL}=\angle{BCL}=90^{\circ}\implies BF \perp LK$ $\blacksquare$
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