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?
"
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,
$\triangle \mathrm{PBC} \sim \triangle \mathrm{PDE}$.
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