20. The "heart" shown in the diagram is formed from an equilateral triangle $A B C$ and two congruent semicircles on $A B$. The two semicircles meet at the point $P$. The point $O$ is the centre of one of the semicircles. On the semicircle with centre $O$, lies a point $X$. The lines $X O$ and $X P$ are extended to meet $A C$ at $Y$ and $Z$ respectively. The lines $X Y$ and $X Z$ are of equal length.
What is $\angle Z X Y$ ?
A $20^{\circ}$
B $25^{\circ}$
C $30^{\circ}$
D $40^{\circ}$
E $45^{\circ}$
Solution
A
Let $\angle Z X Y=x^{\circ}$.
The triangle $P O X$ is isosceles because $O P$ and $O X$ are radii of a semicircle and hence are equal. Therefore, $\angle O P X=$ $\angle O X P=x^{\circ}$.

Since $\angle A P Z$ and $\angle O P X$ are vertically opposite, $\angle A P Z=\angle O P X=x^{\circ}$.

Because it is an angle of an equilateral triangle, $\angle Z A P=60^{\circ}$.

It now follows from the External Angle Theorem applied to the triangle $A Z P$ that $\angle Y Z X=\angle Z A P+\angle A P Z=(60+x)^{\circ}$.

Because $X Y=X Z$, the triangle $X Y Z$ is isosceles and therefore $\angle Z Y X=$ $\angle Y Z X=(60+x)^{\circ}$.

We now apply the fact that the sum of the angles of a triangle is $180^{\circ}$ to the triangle $X Y Z$. This gives
$$
x+(60+x)+(60+x)=180 .
$$

Hence
$$
3 x+120=180 .
$$

It follows that $x=20$ and hence $\angle Z X Y=20^{\circ}$.