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.

Step 4 of 5: Prove the converse

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In plain words

Compare the two pairs of equal radii by projection.

AH\cdot BK=CH\cdot DK\Rightarrow PA=PC

Detailed analysis

Assume the areas are equal. If $PA>PC$ , then the projections on $AC$ give $AH>CH$ ; because $PA=PB$ and $PC=PD$ , the projections on $BD$ give $BK>DK$ . This contradicts $AH\cdot BK=CH\cdot DK$ . The case $PA<PC$ similarly gives the reverse strict inequality. Therefore $PA=PC$ .