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 2 of 5: Reduce the area condition

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In plain words

Replace the original areas by products on the perpendicular diagonals.

[ABP]=[CDP]\iff AH\cdot BK=CH\cdot DK

Detailed analysis

Decompose the two areas at $X$ and use $AC\perp BD$ . In either possible order of $A,X,C$ and $B,X,D$ , the signed-area terms involving $PH$ and $PK$ cancel. The equality is therefore equivalent to $AH\cdot BK=CH\cdot DK$ , where all four quantities are positive distances along the two diagonals.