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

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Let $K$ be the intersection of $AC$ and $BE$ , let $L$ be the intersection of $AC$ and $DF$ ,

[asy] size(10cm); 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,-1.6), E); label(" $L$ ", (0.7,-1.6) ); [/asy]

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

, so  is an isosceles triangle.
Since , so  and  are isosceles triangles. So  is on the angle bisector oof , since  is 
an isosceles trapezoid, so  is also on the perpendicular bisector of . So .

~szhangmath

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

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