23. The points $S, T, U$ lie on the sides of the triangle $P Q R$, as shown, so that $Q S=Q U$ and $R S=R T$.
$\angle T S U=40^{\circ}$. What is $\angle T P U$ ?
A $60^{\circ}$
B $70^{\circ}$
C $80^{\circ}$
D $90^{\circ}$
E $100^{\circ}$
Solution: $\mathbf{E}$
Let $\angle R S T=x^{\circ}$ and $\angle Q S U=y^{\circ}$. As $R S=R T$, the triangle $R S T$ is isosceles and hence $\angle R T S=R S T=x^{\circ}$. Hence, as the angles in triangle $R S T$ add up to $180^{\circ}$, we have that $\angle S R T=(180-2 x)^{\circ}$. Similarly $\angle S Q U=(180-2 y)^{\circ}$. Hence from triangle $P Q R$, we have that $\angle T P U=(180-(180-2 x)-(180-2 y))^{\circ}=(2(x+y)-180)^{\circ}$. Now, as $\angle R S T, \angle Q S U$ and $\angle T S U$ are three angles on a straight line, $x+y+40=180$. Therefore $x+y=140$. We deduce that $\angle T P U=2(140)-180=100^{\circ}$. [Note that although we can find $x+y$, the information given does not enable us to calculate the values of $x$ and $y$ separately.]