13. The diagram shows three triangles which are formed by the five line segments $A C D F, B C G$, $G D E, A B$ and $E F$ so that $A C=B C=C D=G D=D F=E F$. Also $\angle C A B=\angle E F D$.
What is the size, in degrees, of $\angle C A B$ ?
A 40
B 45
C 50
D 55
E 60
Solution
$\mathbf{E}$

Let the size in degrees of $\angle C A B$ be $x$. Since $A C=B C$, triangle $A B C$ is isosceles. Hence $\angle A B C=x^{\circ}$. Since angles in a triangle add to $180^{\circ}$, we have $\angle B C A=(180-2 x)^{\circ}$. Also, since vertically opposite angles are equal, we have $\angle G C D=(180-2 x)^{\circ}$.

The same argument can then be applied to triangle $C G D$ which is isosceles since $C D=G D$. Hence $\angle C D G=(180-2(180-2 x))^{\circ}=(4 x-180)^{\circ}$. Therefore, since vertically opposite angles are equal, $\angle F D E=(4 x-180)^{\circ}$.

The same argument can be applied once more, this time to triangle $F D E$, which is also isosceles since $D F=E F$. This gives $\angle E F D=(180-2(4 x-180))^{\circ}=(540-8 x)^{\circ}$.

However, we are also told that $\angle C A B=\angle E F D$ and hence $x=540-8 x$ or $9 x=540$. This has solution $x=60$ and hence the size, in degrees, of $\angle C A B$ is 60 .