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 = 2C


1 + 2 = 2C (Therefore, CAD = 1 + 2)


2 + 2 = 2C (Therefore, AE bisects CAD)


22 = 2C


2 = C (Alternate angles)


Therefore, AE BC


Hence, proved


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