# $\mathrm{AB}$ 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) $\mathrm{AD}=\mathrm{BE}$"

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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,

$AD=BE$.

Updated on 10-Oct-2022 13:40:54