39th IMO

[sourav das]

Problem 1. In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside
ABCD . Prove that ABCD is a cyclic quadrilateral if and only if the triangles ABP and CDP have equal areas.

[Tahmid Hasan]

the 'if' part.first we consider ABCD is cyclic.then $P$ is the circumcenter.
now,$\triangle APB= \frac{1}{2} BP.AP.sin \angle APB= \frac{1}{2}.R^2.sin \angle APB$
and $\triangle CPD= \frac{1}{2}CP.PD.sin \angle CPD$
$=\frac{1}{2}.R^2.sin 2 \angle CBD$
$=\frac{1}{2}R^2.sin 2 (\angle 90^\circ-\angle ACB)$
$=\frac{1}{2}R^2.sin (180^\circ-\angle APB)$
$=\frac{1}{2}.R^2.sin \angle APB$
$=\triangle APB$
now for the only if part,
for the sake of contradiction we consider the existence of a cyclic quadrilateral $ABCD$ for which the area of $\triangle APB,\triangle CPD$ are not equal.but the 'if' process can be done here and proven $\triangle APB=\triangle CPD$,so a contradiction.

[nayel]

Just a little note: your "if" part is actually the "only if".

[sourav das]

now for the only if part,for the sake of contradiction we consider the existence of a cyclic quadrilateral $ABCD$ for which the area of $\triangle APB,\triangle CPD$ are not equal.but the 'if' process can be done here and proven $\triangle APB=\triangle CPD$,so a contradiction.

Completely wrong. You don't assume any quadrilateral $ABCD$ which satisfy $\triangle APB=\triangle CPD$ ( not cyclic quadrilateral ). Here you actually count only if part twice.