Question 6:
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Answer:
It is given that ∠BAD = ∠EAC
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
In ∆BAC and ∆DAE,
AB = AD (Given)
∠BAC = ∠DAE (Proved above)
AC = AE (Given)
∆BAC ≅ ∆DAE (By SAS congruence rule)
∴ BC = DE (By CPCT)
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