Let ABC be the triangle with the following properties

Question

I am stuck with the following problem:

Can someone point me in the right direction (about problem no. 59)?

Thanks in advance for your time.

Answer 1

Consider the $\triangle ABC $ and $\triangle BDC $. We see from the figure that $\angle BCD =\angle ACB=z+x$. Now, considering the figure , $$z+x+x+ \angle BDC=180^{\circ}=x+x+z+x+y \implies \angle BDC=x+y.$$ Hence $\triangle ABC $ and $\triangle BDC $ are similar. So, $\frac{BC}{AC}=\frac{DC}{BC}\implies \frac{BC}{9}=\frac{4}{BC} \implies BC=6.$

The perimeter of $\triangle BEC$ being $13$, $BE+EC+6=13 \implies BE=EC=\frac{13-6}{2}=3.5$