In a convex quadrilateral $ABCD$ we have $ADC = 90^{\circ}$. Let $E$ and $F$ be the projections of $B$ onto the lines $AD$ and $AC$, respectively. Assume that $F$ lies between $A$ and $C$, that $A$ lies between $D$ and $E$, and that the line $EF$ passes through the midpoint of the segment $BD$. Prove that the quadrilateral $ABCD$ is cyclic.
Problem
Source: Baltic Way 2007
Tags: geometry, circumcircle, geometry proposed
AK1024
30.11.2010 22:04
Let $M$ be the midpoint of $CD$. Furthermore, let $EF$ meet $CD$ at $P$ and $EF$ meet $AB$ at $Q$ (shown in the diagram). Since both $BE$ and $CD$ are parallel to $AD$ they are parallel, which means $\angle BEP=\angle EPD$. Consider triangle $BDE$'s circumcircle. It has diameter $BD$ since $\angle BED=90^{\circ}$. Hence $M$ is the circumcentre and so $BM=EM=DM$. So $\triangle MEB$ is isosceles $\implies \angle BEM=\angle EBM=\angle PDB\implies BEDP$ is cyclic. But then, using the fact that $AFBE$ is cyclic, $\angle BDC=\angle BDP=\angle BEP=\angle BEF=\angle BAF=\angle BAC$. Thus $ABCD$ is cyclic. I was going to use $Q$ but I changed my mind
BSJL
18.02.2014 17:24
There is a shorter solution! Let $ G $ be the projection of $ B $ onto the line $ CD $, then it's clear that $ E,F,G $ are collinear because the line $ EG $ is also pass the midpoint of segment $ BD $. So, we conclude $ B $ is on the circumcircle of the triangle $ ADC $ by Simson Theorem.
MMEEvN
11.05.2014 16:07
Even shorter $N$ the midpoint of $AB$ . $\Rightarrow BN=EN$ and $MN||ED \Rightarrow MN \perp EB$ $\Rightarrow MB=EB \Rightarrow \angle CAB=\angle MEB=\angle MBE=\angle BDC $
Mikasa
11.05.2014 17:08
Let $M$ be the midpoint of $BD$. Now since $\angle BED=90^{\circ}, M$ must be the center of the circle $BED$. So $ME=MB$. Now $\angle BDC=90^{\circ}-\angle EDB=\angle EBD=\angle EBM=\angle BEM=\angle BEF$. But since $\angle BEA=\angle BFA=90^{\circ}$ the points $A,E,B,F$ are concyclic. Then $\angle BEF=\angle BAF=\angle BAC$. Thus $\angle BDC=\angle BAC$ and we are done.
Korapus5732
02.06.2015 11:21
MMEEvN wrote: Even shorter $N$ the midpoint of $AB$ . $\Rightarrow BN=EN$ and $MN||ED \Rightarrow MN \perp EB$ $\Rightarrow MB=EB \Rightarrow \angle CAB=\angle MEB=\angle MBE=\angle BDC $ Nice one You could also try the Simpson's Line of $B$ w.r.t $ADC$
samithayohan
02.06.2015 12:17
Yes Korapus5732.This is my solution using Simpsons line.G is the feet of the perpendicular from B to DC.M is the midpoint of BD "EBGD is a rectangle $\implies$ E,M,G collinear $\implies$ E,F,G collinear $\implies$ ABCD cyclic(From the converse of simpsons line)" This might even be shorter than MMEEvNs solution