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

(A) To prove: $\text{arc } CXA + \text{arc } DZB = \text{arc } AYD + \text{arc } BWC = \text{semicircle}$.
Construction: Join $AC, AD, BD$ and $BC$.
Proof: Let the chords $AB$ and $CD$ intersect at point $O$ at right angles. Thus,$\angle AOC = \angle COB = \angle BOD = \angle DOA = 90^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Let the measure of the arcs be denoted by their corresponding central angles.
In $\triangle AOC$,$\angle OAC + \angle OCA = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Since the angle subtended by an arc at the circumference is half the angle at the center,we have:
$\text{arc } CXA = 2 \angle CBA$ and $\text{arc } DZB = 2 \angle BCD$.
In $\triangle OBC$,$\angle OBC + \angle OCB = 90^{\circ}$.
Thus,$\frac{1}{2} (\text{arc } CXA) + \frac{1}{2} (\text{arc } DZB) = \angle CBA + \angle BCD = 90^{\circ}$.
Therefore,$\text{arc } CXA + \text{arc } DZB = 180^{\circ}$,which is a semicircle.
Similarly,for the other pair,$\text{arc } AYD + \text{arc } BWC = 180^{\circ}$,which is also a semicircle.
Hence,$\text{arc } CXA + \text{arc } DZB = \text{arc } AYD + \text{arc } BWC = \text{semicircle}$.