Let the intersection points of the diagonals is denoted by P,

then, because, angle-APD = pi/2, so the measure of the sum of the arcs (AD)+(BC)=pi,

similarly, the measure of the sum of the arcs (DC)+(BA)=pi.

So, respectively,

area-of-triangle-AOD + area-of-triangle-DOC =

= sin(angle AOD) + sin(angle DOC) =

= the area in the quadrilateral on the right-hand-side

is the same as

= sin(pi - angle AOD) + sin(pi - angle DOC) =

= sin(angle COB) + sin(angle BOA) =

= area-of-triangle-COB + area-of-triangle-BOA

= the area in the quadrilateral on the left-hand-side,

where without loss of generality, the length of the radius is assumed to be sqrt(2).

Another approach:

Let r be the radius of the circle.

1. Draw diameter BE, intersecting (CA) at point P.

2. BDE is a right triangle, angle BED = angle EPA

3. sin(angle BED) = BD/2r = sin(angle EPA)

4. A1 = area of quadrilateral ABCD = (1/2) * (BD) * (CA)

5. A2 = area of quadrilateral ABCO = (1/2) * r * (CA) * sin(angle EPA)

6. A1/A2 = {(1/2) * (BD) * (CA)} / {(1/2) * r * (CA) * sin(angle EPA)}

= {(1/2) * (BD) * (CA)} / {(1/2) * r * (CA) * BD/2r)} = 2

QED

Thank you for introducing this interesting geometric problem.

Steftcho P. Dokov Thank you for your time.

George McIlvaine I appreciate you for sharing your solution! Interesting approach and a very accessible solution!