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$ \mathrm{AB} $ is a line segment and $ \mathrm{P} $ is its mid-point. $ \mathrm{D} $ and $ \mathrm{E} $ are points on the same side of $ \mathrm{AB} $ such that $ \angle \mathrm{BAD}=\angle \mathrm{ABE} $ and $ \angle \mathrm{EPA}=\angle \mathrm{DPB} $ (see Fig. 7.22). Show that
(i) $ \triangle \mathrm{DAP} \cong \triangle \mathrm{EBP} $
(ii) $ \mathrm{AD}=\mathrm{BE} $
"
Given:
$AB$ is a line segment and $P$ is its mid-point.
$D$ and $E$ are points on the same side of $AB$ such that $\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$.
To do:
We have to show that
(i) $\triangle DAP \cong \triangle EBP$
(ii) $AD=BE$.
Solution:
(i) Let us add $\angle DPE$ on both sides of $\angle EPA=\angle DPB$ we get,
$\angle EPA+\angle DPE=\angle DPB+\angle DPE$
This implies,
$\angle DPA=\angle EPB$
Now, let us consider $\triangle DAP$ and $EBP$
We have,
$DPA=EPB$
We also know that $P$ is the midpoint of the line segment $AB$
This implies,
$AP=BP$
Since $\angle BAD=\angle ABE$
Therefore,
By Angle-Side-Angle:
If two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent.
We get,
$\triangle DAP \cong \triangle EBP$
(ii) We know that,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.
This implies,
$AD=BE$.
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