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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AcademicMathematicsNCERTClass 10

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,

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

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

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