AB is a line segment and P is its midpoint. D and E are points on the

1.	AB is a line segment and P is its 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.
Answer» Data : 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 : (i) ∆DAP ≅ ∆EBP (ii) AD = BE Proof : (i) In ∆DAP and ∆EBP, AP = BP (∵ P is the mid-point of AB) ∠BAD = ∠ABE (Data) ∠APD = ∠BPE ∵ (∠EPA = ∠DPB Adding ∠EPD to both sides) ∠EPA + ∠EPD = ∠DPB + ∠EPD ∴ ∠APD = -BPE. Now, Angle, Side, Angle postulate. ∴ ∆DAP ≅ ∆EBP . (ii) As it is ∆DAP ≅ ∆EBP , Three sides and three angles are equal to each other. ∴ AD = BE.

AB is a line segment and P is its 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.

Answer»

Data : 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 :

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Proof : (i) In ∆DAP and ∆EBP,

AP = BP (∵ P is the mid-point of AB)

∠BAD = ∠ABE (Data)

∠APD = ∠BPE

∵ (∠EPA = ∠DPB

Adding ∠EPD to both sides)

∠EPA + ∠EPD

= ∠DPB + ∠EPD

∴ ∠APD = -BPE.

Now, Angle, Side, Angle postulate.

∴ ∆DAP ≅ ∆EBP .

(ii) As it is ∆DAP ≅ ∆EBP ,

Three sides and three angles are equal to each other.

∴ AD = BE.