Let $E$ be intersection of the diagonals of convex quadrilateral $ABCD$. It is given that $m(\widehat{EDC}) = m(\widehat{DEC})=m(\widehat{BAD})$. If $F$ is a point on $[BC]$ such that $m(\widehat{BAF}) + m(\widehat{EBF})=m(\widehat{BFE})$, show that $A$, $B$, $F$, $D$ are concyclic.
Problem
Source: Turkey TST 2013 - Day 3 - P1
Tags: geometry, circumcircle, geometry proposed
MMEEvN
02.04.2013 22:30
It is easy to observe that the point $F$ is unique when $A,B,C,D$ are fixed . Hence we may try to construct such a point $F$.In other words it is enough to show that the intersection of the circumcircle of $\Delta ABD$ and $CB$ satisfies the properties of point $F$.Let the intersection be $X$. $\Rightarrow \angle DXC=\angle DAB =\angle DEC \Rightarrow XEDC$ is cyclic and $\angle XBD = \angle XAD$ and $\angle BAX+\angle XAD=\angle BAD =\angle EDC= \angle CDX+\angle XDE \Rightarrow \angle XDC= \angle CBD \Rightarrow \angle CEX=\angle CBE$ Easy angle chasing gives $\angle ACB = \angle XAB \Rightarrow \angle XBE+\angle BAX =\angle XEC+\angle XCE =\angle BXE$ as desired
Konigsberg
19.05.2014 17:16
Do you mean $\angle XDC = \angle CBD$?
Sardor
17.12.2014 17:13
We must prove that if $ F=BC \cap (ABD) $, then $ \angle BAF + \angle EBF=\angle BFE $ ( because if $ A,B,C,D $ are fixed $ \implies F $ is also fixed).We have from conditions of the problem, $ CD $ is tangent to $ (ABD) $ and $ CD=CE $, and then $ CE^2=CD^2=CF \cdot CB \implies CE $ is tangent to $ (EBF) $.We have $ \angle BAD=\angle CDE=\angle CED=\angle BEA \implies BA $ is tangent to $ (AED) $.Then $ \angle CDF=\angle DAF=\angle DBF=\angle EBF=\angle CEF $, so $ CDEF $ is cyclic.Then we have easily $ \angle BAF + \angle EBF=\angle BFE $.So we are done.