In convex pentagon $ABCDE$, the sides $AE,BC$ are parallel and $\angle ADE=\angle BDC$. The diagonals $AC$ and $BE$ intersect at $P$. Prove that $\angle EAD=\angle BDP$ and $\angle CBD=\angle ADP$.
Problem
Source: Baltic Way 1998
Tags: geometry proposed, geometry
Tiger100
28.06.2009 22:52
Could you clarify what you mean with diameters? is this pentagon cyclic?
behdad.math.math
29.06.2009 11:55
No.AC and BE are lines.
WakeUp
11.01.2011 20:25
Tiger100's confusion was due to your usage of the word 'diameter', instead of diagonal: behdad.math.math wrote: In convex pentagon ABCDE, two sides AE, BC are parallel. <ADE = <BDC two diameters AC, BE intersect each other on P. prove that <EAD=<BDP and <CBD=<ADP.
oneplusone
13.01.2011 15:38
Let the line passing through $B$ parallel to $DE$ intersect $DP$ at $F$. Then $FBCP\sim DEAP$, so $\angle CFB=\angle ADE=\angle BDC$, so $FBCD$ is cyclic. Thus $\angle DBC=\angle DFC=\angle FDA$. Similarly the other one.