Question

If two chords $AB$ and $CD$ of a circle $AYDZBWCX$ intersect at right angles prove that $arc\left(CXA\right)+arc\left(DZB\right)=arc\left(AYD\right)+arc\left(BWC\right)=arc\left(\text{semicircle}\right)$

Answer 1

A chord divides the circumference of a circle in a way that their mean is the circumference of the semicircle.

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

And for mutually perpendicular chords the difference in the length of arcs cut by chords are the same.

So, by using the above statement we can say that,

$BWC-DZB=CXA-AYDCXA+DZB=AYD+BWC\dots \left(i\right)$

And they all sum up to give the circumference of the circle.

$\begin{array}{cc}CWB+BZD+CXA+DYA& =2\pi r\\ 2\left(BWC+AYD\right)=2\pi r\text{or}2\left(CXA+DZB\right)& =2\pi r\text{by using}\left(i\right)\\ BWC+AYD=\pi r\text{or}CXA+DZB& =2\pi r\end{array}$

$\pi r$ is nothing but the length of the semicircle. Hence proved.