In the given figure,$BD$ and $CE$ intersect each other at the point $P$. Is $\triangle PBC \sim \triangle PDE$ ? Why?

(A) Yes,$\triangle PBC \sim \triangle PDE$.
In $\triangle PBC$ and $\triangle PDE$:
$\angle BPC = \angle EPD$ [Vertically opposite angles]
Now,calculating the ratios of the sides including the angles:
$\frac{PB}{PD} = \frac{5 \text{ cm}}{10 \text{ cm}} = \frac{1}{2}$ ......$(i)$
$\frac{PC}{PE} = \frac{6 \text{ cm}}{12 \text{ cm}} = \frac{1}{2}$ ......$(ii)$
From equations $(i)$ and $(ii)$:
$\frac{PB}{PD} = \frac{PC}{PE}$
Since one angle of $\triangle PBC$ is equal to one angle of $\triangle PDE$ and the sides including these angles are proportional,by the $SAS$ (Side-Angle-Side) similarity criterion,the triangles are similar.
Hence,$\triangle PBC \sim \triangle PDE$.