In Fig. 9.38, AE bisects ∠CAD and ∠B =∠C. Prove that AE||BC.
Given,
AE bisects ∠CAD
∠B = ∠C
In
∠CAD = ∠B + ∠C
∠CAD = ∠C + ∠C
∠CAD = 2∠C
∠1 + ∠2 = 2∠C (Therefore, ∠CAD = ∠1 + ∠2)
∠2 + ∠2 = 2∠C (Therefore, AE bisects ∠CAD)
2∠2 = 2∠C
∠2 = ∠C (Alternate angles)
Therefore, AE ‖ BC
Hence, proved