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 3 of 5: Prove the forward implication

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In plain words

A cyclic quadrilateral has one common center.

ABCD\text{ cyclic}\Rightarrow PA=PB=PC=PD

Detailed analysis

If $ABCD$ is cyclic, the perpendicular bisectors of the four sides meet at the circumcenter. Thus $P$ is the circumcenter, so $PA=PB=PC=PD$ . The feet $H$ and $K$ are then the midpoints of both diagonals, hence $AH=CH$ and $BK=DK$ , which gives equal areas by the equivalence above.