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2004 IMO Problems/Problem 5

Problem

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.\]

Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP.$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Assume $ABCD$ is cyclic, let $K$ be the intersection of $AC$ and $BE$, let $L$ be the intersection of $AC$ and $DF$,

[asy] size(6cm); draw(circle((0,0),7.07)); draw((-3.7,-6)-- (3.7,-6)); draw((-6.8,-2)-- (6.8,-2)); draw((-5,5)-- (5,5)); draw((-5,5)-- (-3.7,-6)); draw((-5,5)-- (3.7,-6)); draw((-5,5)-- (-6.8,-2)); draw((-5,5)-- (6.8,-2)); draw((5,5)-- (-3.7,-6)); draw((5,5)-- (3.7,-6)); draw((5,5)-- (-6.8,-2)); draw((5,5)-- (6.8,-2)); draw((-3.7,-6)-- (-6.8,-2)); draw((-3.7,-6)-- (6.8,-2)); draw((3.7,-6)-- (-6.8,-2)); draw((3.7,-6)-- (6.8,-2)); label("$A$", (-6.8,-2), SW); label("$B$", (-3.7,-6), SW); label("$F$", (3.7,-6), SE); label("$C$", (6.8,-2), E); label("$E$", (5,5), E); label("$D$", (-5,5), W); label("$P$", (0,-1.3), N); label("$K$", (-1.6,-1.5), E); label("$L$", (0.8,-1.5) ); [/asy]

$\angle PBC=\angle DBA$, so $AD=CE$, and $DE//AC$. $\angle PDC=\angle BDA$, so $AB=CF$, and $AC//BF$. 

$\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL$, so $\triangle PKL$ is an isosceles triangle.
Since $AC//BF$, so $\triangle PBF$ and $\triangle PDE$ are isosceles triangles. So $P$ is on the perpendicular bisector of $BF$, since $ABFC$ is 
an isosceles trapezoid, so $P$ is also on the perpendicular bisector of $AC$. So $PA=PC$.


~szhangmath

See Also

2004 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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