Question

AB is a line segment and P is it's midpoint.D and E are points on the same side of AB such that angle BAD=angle ABE and angle EPA=angle DPA .show that (i) triangle DAP is congruence to triangle EBP (ii)AD=BE​

Answer 1

Answer:

AB is a line segment and P is it's midpoint.D and E are points on the same side of AB such that angle BAD=angle ABE and angle EPA=angle DPA .show that (i) triangle DAP is congruence to triangle EBP (ii)AD=BE

Answer 2

Correct Question?

AB is a line segment and P is it's midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that (i) ∆DAP ≅ ∆EBP (ii) AD = BE

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Given :

• AB is a line segment and P is it's midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB.

To Prove :

• ∆DAP ≅ ∆EBP
• AD = BE

Proof :

Since, P is the midpoint of the line segment AB$.$

So ,

→ AP = PB

→ ∠DAP = ∠EBP

→ ∠EPA = ∠DPB

Add ∠EPD to both sides (∵ ∠EPD is common)

⇒ ∠EPA + ∠EPD = ∠DPB + ∠EPD

⇒ ∠DPA = ∠EPD .....1

In ∆DAP and ∆EBP

⪼ ∠DPA = ∠EPD [From equation (1)]

⪼ AP = BP [Given]

⪼ ∠DAP = ∠EBP [Given]

∴ ∆DAP ≅ ∆EBP [By ASA-congruence]

∴ AD = BE [By CPCT]

Hence Proved !