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In Fig. 7.21, $ \mathrm{AC}=\mathrm{AE}, \mathrm{AB}=\mathrm{AD} $ and $ \angle \mathrm{BAD}=\angle \mathrm{EAC} $. Show that $ \mathrm{BC}=\mathrm{DE} $.
"
Given:
$AC=AE,AB=AD$ and $\angle BAD=\angle EAC$.
To do:
We have to show that $BC=DE$.
Solution:
Let us add $\angle DAC$ on both sides of $\angle BAD=\angle EAC$ we get,
$\angle BAD+\angle DAC=\angle EAC+\angle DAC$
This implies,
$\angle BAC=\angle EAD$
We know that,
According to Rule of Side-Angle-Side Congruence:
Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Given,
$AC=AE$ and $AB=AD$
We also have,
$\angle BAC=\angle EAD$
Therefore,
$\angle ABC \cong \angle ADE$.
We also know,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding sides must be equal.
Therefore,
$BC=DE$.
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