In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
It is given that EPA = DPB
∴∠EPA +∠ DPE = ∠DPB + ∠DPE
∴ ∠DPA = ∠EPB
In ∆DAP and ∆EBP,
∠DAP = ∠EBP (Given)
AP = BP (P is mid-point of AB)
∠DPA = ∠EPB (From above)
∠∆DAP ∠∆EBP (ASA congruence rule)
∴ AD = BE (By CPCT)