In Fig. 10.22, th

Given,

The sides BA and CA have been produced, such that:

And, CA = AE

We have to prove that,

DE BC

Consider and , we have

BA = AD and CA = AE (Given)

BAC = DAE (Vertically opposite angle)

So, by SAS congruence rule we have: Therefore, BC = DE and

DEA = BCA,

EDA = CBA (By c.p.c.t)

Now, DE and BC are two lines intersected by a transversal DB such that,

DEA = BCA,

i.e., Alternate angles are equal

Therefore, DE BC

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