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 (see Fig. 7.22). Show that
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE

 

 

 

 

 

 

 

 

Given: P is its mid-point of AB, ∠BAD = ∠ABE and ∠EPA = ∠DPB

To Prove: i) ΔDAP ≅ ΔEBP and (ii) AD = BE.

Proof: i) It is given that ∠EPA = ∠DPB

∠EPA + ∠DPE = ∠DPB + ∠DPE (∠DPE is common)

∴ ∠DPA = ∠EPB

In ΔDAP and ΔEBP,

∠DAP = ∠EBP (Given)

AP = BP (P is mid - point of AB)

∠DPA = ∠EPB (Proven above)

∴ ΔDAP ≅ ΔEBP (ASA congruence rule)

ii) Since, ΔDAP ≅ ΔEBP 

∴ AD = BE (By CPCT)

Hence, proved.