12. A circle touches the sides of triangle $P Q R$ at the points $S, T$ and $U$ as shown. Also $\angle P Q R=\alpha^{\circ}, \angle P R Q=\beta^{\circ}$ and $\angle T S U=\gamma^{\circ}$.
Which of the following gives $\gamma$ in terms of $\alpha$ and $\beta$ ?
A $\frac{1}{2}(\alpha+\beta)$
B $180-\frac{1}{2}(\alpha+\beta)$
C $180-(\alpha+\beta)$
D $\alpha+\beta$
E $\frac{1}{3}(\alpha+\beta)$
Solution
$\mathbf{A}$
METHOD 1
From the given fact that the circle touches the sides of the triangle, we deduce that the lines $P Q, Q R$ and $R P$ are tangents to the circle. The two tangents from a given point to a circle are of equal length. In particular, $Q S=Q U$. Therefore the triangle $Q S U$ is isosceles, and hence $\angle Q S U=\angle Q U S$.
Because the sum of the angles in a triangle is $180^{\circ}$, from the triangle $Q S U$ we deduce that $\alpha^{\circ}+\angle Q S U+\angle Q U S=180^{\circ}$.

Since $\angle Q S U=\angle Q U S$, it follows that $\alpha^{\circ}+2 \angle Q S U=180^{\circ}$. We can rearrange this equation to give
$$
\angle Q S U=\frac{1}{2}(180-\alpha)^{\circ} .
$$

In a similar way, we have that the tangents $R S$ and $R T$ are of equal length. Hence the triangle $R T S$ is isosceles and therefore
$$
\angle R S T=\frac{1}{2}(180-\beta)^{\circ} .
$$

Because $Q S R$ is a straight line, the angles at $S$ on this line have sum $180^{\circ}$, that is,
$$
\angle Q S U+\angle R S T+\gamma^{\circ}=180^{\circ} \text {. }
$$

Substituting in this equation the expressions for $\angle Q S U$ and $\angle R S T$ we have already found, we deduce that
$$
\frac{1}{2}(180-\alpha)^{\circ}+\frac{1}{2}(180-\beta)^{\circ}+\gamma^{\circ}=180^{\circ} .
$$

We can rearrange this equation as
$$
180^{\circ}-\frac{1}{2}(\alpha+\beta)^{\circ}+\gamma^{\circ}=180^{\circ},
$$
from which it follows that
$$
\gamma=\frac{1}{2}(\alpha+\beta) .
$$

Method 2
As in method $1, P Q$ and $P T$ are tangents to the circle.
By the Alternate Angle Theorem, $\angle P U T=\angle P T U=\gamma^{\circ}$. Therefore because the angles in the triangle $P U T$ have sum $180^{\circ}$, we have $\angle U P T+\gamma^{\circ}+\gamma^{\circ}=180^{\circ}$, and therefore
$$
\angle U P T=180^{\circ}-2 \gamma^{\circ} \text {. }
$$

Therefore, as the angles in the triangle $P Q R$ have sum $180^{\circ}$,
$$
\alpha^{\circ}+\beta^{\circ}+\left(180^{\circ}-2 \gamma^{\circ}\right)=180^{\circ} \text {. }
$$

We can rearrange this last equation to give
$$
2 \gamma^{\circ}=\alpha^{\circ}+\beta^{\circ},
$$
and it follows that
$$
\gamma=\frac{1}{2}(\alpha+\beta) .
$$

For INVESTIGATION
12.1 In Method 1 we used the fact that the two tangents from a given point to a circle are of equal length. Find a proof of this (that is, either devise your own proof, look in a book or on the internet, or ask your teacher).
12.2 In Method 2 we have used the Alternate Segment Theorem which says that the angle between a tangent to a circle and a chord is equal to the angle subtended by the chord at the circumference of the circle. Find a proof of this theorem.
12.3 Here we consider a third method.
We let $O$ be the centre of the circle. In the figure we have added the lines joining $O$ to the points $Q, S$, and $U$. We let $V$ be the point where $Q O$ meets $U S$.
Now proceed as follows.
(i) Prove that the triangles $Q O S$ and $Q O U$ are congruent.
(ii) Deduce that $\angle V Q S=\angle V Q U=\frac{1}{2} \alpha^{\circ}$.
(iii) Prove that the triangles $Q V S$ and $Q V U$ are congruent.
(iv) Deduce that $\angle Q V S=90^{\circ}$.
(v) Deduce that $\angle U S O=\frac{1}{2} \alpha^{\circ}$.
(vi) Show that, similarly, $\angle T S O=\frac{1}{2} \beta^{\circ}$.
(vii) Conclude that $\gamma=\frac{1}{2}(\alpha+\beta)$.