# In figure below, $\mathrm{BD}$ and $\mathrm{CE}$ intersect each other at the point $\mathrm{P}$. Is $\triangle \mathrm{PBC} \sim \triangle \mathrm{PDE}$ ? Why?"

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Given:

$\mathrm{BD}$ and $\mathrm{CE}$ intersect each other at the point $\mathrm{P}$.

To do:

We have to find whether $\triangle \mathrm{PBC} \sim \triangle \mathrm{PDE}$.

Solution:

From the figure,

$\angle B P C=\angle E P D$             (Vertically opposite angles)

$\frac{P B}{P D}=\frac{5}{10}=\frac{1}{2}$...........(i)

$\frac{P C}{P E}=\frac{6}{12}$

$=\frac{1}{2}$.............(ii)

From (i) and (ii), we get,

$\frac{P B}{P D}=\frac{P C}{P E}$

Here,

One of the angles of $\triangle \mathrm{PBC}$ is equal to one of the angles of $\triangle \mathrm{PDE}$ and the sides including these angles are proportional.

Therefore, both the triangles are similar.

Hence, by SAS similarity,

$\triangle \mathrm{PBC} \sim \triangle \mathrm{PDE}$.

Updated on 10-Oct-2022 13:27:57