Question

If two chords $AB$ and $CD$ of a circle $AYDZBWCX$ intersect at right-angle, Prove that area $CXA$ + arc $DZB$ is semicircle.

Answer 1

A chord divides the circle in segments such a way that their mean is area of semicircle.

The difference in the area of segments is inversely proportional to length of chord.

And for mutually perpendicular chords the difference of the area of segments cut by chords are same

so, $CWB-BZD=CXA-DYACXA+DZB=AYD+BWC$

And they all sum up to give Area of circle

$CWB+BZD+CXA+DYA=\pi r^{2}CWB+DYA=\frac{\pi r^2}{2}$(semicircle).