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 5 of 5: Conclude cyclicity

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In plain words

All four vertices share one center and one radius.

PA=PB=PC=PD

Detailed analysis

From the definition of $P$ , $PA=PB$ and $PC=PD$ ; together with $PA=PC$ , all four vertices are on the circle centered at $P$ . Hence $ABCD$ is cyclic.