AB is a line segment and P is its mid-point. D and E are points o… - Vidyadip

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 $\angle EPA = \angle DPB.$
Show that:
$i. \triangle DAP \cong \triangle EBP$
$ii. AD = BE$

✓

Answer

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 prove:
$i. \text{DDAP} \cong \text{DEBP}$
$ii. AD = BE$
Proof :$(ii)$
$\angle EPA = \angle DPB ...[$Given$]$
$\angle EPA + \angle EPD = \angle EPD + \angle DPB ...[$Adding $\angle EPD$ to both sides$]$
$∠APD = ∠BPE ...(1)$
In $DDAP$ and $DEBP$
$\angle DAP = \angle EBP ...[$Given$]$
$AP = BP ...[$As $P$ is the mid$-p$oint of the line $AB]$
$\angle APD = \angle BPE ...[$From $(1)]$
$\therefore DDAP \cong DEBP$ proved$ ...[\text{ASA}$ property$] ...(2)$
$(i)$ As $DDAP \cong DEBP ...[$From $(2)]$
$\therefore AD = BE ... [\text{c.p.c.t.}]$

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