AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that: 1. DAP EBP 2. AD = 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 BAD = ABE and EPA = DPB. To prove: 1. DDAP DEBP 2. AD = BE Proof :(ii) EPA = DPB ...[Given] EPA + EPD = EPD + DPB ...[Adding EPD to both sides] ∠APD = ∠BPE ...(1) In DDAP and DEBP DAP = EBP ...[Given] AP = BP ...[As P is the mid-point of the line AB] APD = BPE ...[From (1)] DDAP DEBP proved ...[ASA property] ...(2) (i) As DDAP DEBP ...[From (2)] AD = BE ...[c.p.c.t.]

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

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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:

$\triangle$DAP $\cong$ $\triangle$EBP
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:

DDAP $\cong$ DEBP
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-point of the line AB]
$\angle$APD = $\angle$BPE ...[From (1)]
$\therefore$ DDAP $\cong$ DEBP proved ...[ASA property] ...(2)
(i) As DDAP $\cong$ DEBP ...[From (2)]
$\therefore$ AD = BE ...[c.p.c.t.]