$ l $ and $ m $ are two parallel lines intersected by another pair of parallel lines $ p $ and $ q $ (see Fig. 7.19). Show that $ \triangle \mathrm{ABC} \cong \triangle \mathrm{CDA} $.
"

Given:

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$.

To do:

We have to show that $\triangle ABC\cong \triangle CDA$.

Solution:

Let us consider $\triangle ABC$ and $\triangle CDA$,

We know that,

When the lines intersected by the transversal are parallel, alternate interior angles are equal.

This implies,

$\angle BCA=\angle DAC$ and $BAC=\angle DCA$

Since $AC$ and $CA$ is the common side of the triangles,

We get,

$AC=CA$

Therefore,

From ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.

We get,

$\triangle ABC \cong\triangle CDA$.

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Updated on: 10-Oct-2022

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