Question

In figure, $ \mathrm{BD}$ and $ \mathrm{CE}$ intersect each other at the point $ \mathrm{P}.$ Is $ \mathrm{\Delta PBC} ~ \mathrm{\Delta PDE}$? Why?

n

• A. True
• B. False

Answer 1

The correct option is A True

Find whether $\mathrm{\Delta PBC}~\mathrm{\Delta PDE}$ is true or false.

In $\mathrm{\Delta PBC}\&\mathrm{\Delta PDE},$

$$\begin{gathered} \frac{\mathrm{BP}}{\mathrm{PD}}=\frac{5}{10}=\frac{1}{2}-\left(1\right) \\ \frac{\mathrm{PC}}{\mathrm{PE}}=\frac{6}{12}=\frac{1}{2}-\left(2\right) \end{gathered}$$

From (1) and (2)

$\frac{\mathrm{BP}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PE}}$

$\angle \mathrm{CPB}=\angle \mathrm{EPD}$ (vertically opposite angles)

$\therefore ∆\mathrm{PBC}~∆\mathrm{PDE}$ (by SAS similarity)

Hence the given statement is true.