Question

$AB$ 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 $\angle EPA=\angle DPB$. Then:

• A. $AB=AD$
• B. $△DAP\cong △EBP$
• C. $AD=BE$
• D. $\angle DPA=\angle EPB$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $△DAP\cong △EBP$
C $AD=BE$
D $\angle DPA=\angle EPB$

$\angle EPA=\angle DPB$ (Given)
$\angle EPA$ $+\angle DPE$ $=\angle DPB$ $+\angle DPE$
$⟹$ $\angle APD=\angle BPE$........... $\left(1\right)$.
Now, in $△DAP$ and $△EBP$,
$AP=PB$ .....($\because P$ is the mid point of $AB$)
$\angle PAD=\angle PBE$ .....(Given)
$\angle APD=\angle BPE$ .....(From $1$).

Therefore $△DAP\cong △EPB$ (By $ASA$ congruency criterion)
Thus, $AD=BE$ (By $CPCT$ rule).

Hence, options $B$, $C$ and $D$ are correct.