In a convex quadrilateral $ABCD$ diagonals intersect at $E$ and $BE = \sqrt{2}\cdot ED, \: \angle BEC = 45^{\circ}.$ Let $F$ be the foot of the perpendicular from $A$ to $BC$ and $P$ be the second intersection point of the circumcircle of triangle $BFD$ and line segment $DC$. Find $\angle APD$.
Problem
Source: Turkey JBMO Team Selection Test 2013, P7
Tags: geometry, circumcircle, geometry proposed
sunken rock
01.06.2013 14:51
Let $Q$ be the foot of perpendicular from $B$ onto $AC$; from the given segments relation, $QE=DE$, since $\triangle BEQ$ is right-angled and isosceles. Hence $\angle DQE=\angle QDE=22.5^\circ$. In $\triangle ABC, AF, BQ$ are altitudes, so $CB\cdot CF=CQ\cdot AC$, while p.o.p. $C$ w.r.t. $\odot (BFD)$ gives $CB\cdot CF=CP\cdot CD$, consequently $AQPD$ is cyclic, and $\angle APD=\angle AQD=22.5^\circ$. Nice problem! Best regards, sunken rock
mathuz
02.06.2013 16:56
it's nice problem! My solution same with sunken rock's solution.
ArseneLupin
02.06.2013 18:38
sunken rock wrote: Let $Q$ be the foot of perpendicular from $B$ onto $AC$; from the given segments relation, $QE=DE$, since $\triangle BEQ$ is right-angled and isosceles. Hence $\angle DQE=\angle QDE=22.5^\circ$. In $\triangle ABC, AF, BQ$ are altitudes, so $CB\cdot CF=CQ\cdot AC$, while p.o.p. $C$ w.r.t. $\odot (BFD)$ gives $CB\cdot CF=CP\cdot CD$, consequently $AQPD$ is cyclic, and $\angle APD=\angle AQD=22.5^\circ$. Nice problem! Best regards, sunken rock I think that $\angle APD+\angle AQD=180^\circ$. So $\angle APD=157.5^\circ$??
mathuz
02.06.2013 20:50
but $ \angle BEC=45\circ.$