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 1 of 5: Introduce the diagonal intersection

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In plain words

Use the perpendicular diagonals and the two perpendicular bisectors.

X=AC\cap BD

Detailed analysis

Let $X$ be the intersection of $AC$ and $BD$ . Let $H,K$ be the feet of the perpendiculars from $P$ to $AC,BD$ , respectively. Since $P$ lies on the perpendicular bisectors of $AB$ and $CD$ , we have $PA=PB$ and $PC=PD$ .