2015-I-13

$\begin{array}{ll}
AE = BF & \text{(given)} \\
\angle ABE = \angle BCF = 90^\circ & \text{(property of square)} \\
AB = BC & \text{(property of square)}
\end{array}$

Therefore, $\Delta ABE \cong \Delta BCF \text{ (RHS)}$.

$\begin{array}{ll}
\angle FBE = \angle EAB & \text{(corr. $\angle$s, $\cong \Delta$s)}
\end{array}$

In $\Delta BGE$,

$\begin{array}{ll}
\angle BGE = 180^\circ – \angle GBE – \angle GEB & \text{($\angle$ sum of $\Delta$)} \\
\angle BGE = 180^\circ – \angle EAB – \angle AEB & \text{(proved)} \\
\angle BGE = \angle ABE & \text{($\angle$ sum of $\Delta$)} \\
\angle BGE = 90^\circ & \text{(property of square)}
\end{array}$

Therefore, $\Delta BGE$ is a right-angled triangle.

$\begin{array}{ll}
BE = CF & \text{(corr. sides, $\cong\Delta$s)} \\
BE = 15 \text{ cm}
\end{array}$

Then by applying Pythagoras Theorem $\Delta BGE$, we have

$\begin{array}{rcl}
BG^2 & = & BE^2 – EG^2 \\
BG & = & \sqrt{15^2 – 9^2} \\
BE & = & 12 \text{ cm}
\end{array}$