In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Answer
Given,
AC = AE, AB = AD and ∠BAD = ∠EAC
To show,
BC = DE
Proof,
∠BAD = ∠EAC (Adding ∠DAC both sides)
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
In ΔABC and ΔADE,
AC = AE (Given)
∠BAC = ∠EAD
AB = AD (Given)
Therefore, ΔABC ≅ ΔADE by SAS congruence condition.
BC = DE by CPCT.
