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In Fig. 10.22, the sides BA and CA have been produced such that: BA = AD and CA = AE.

Prove that segment DE || BC.

Fig. 10.22

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Given that, the sides BA and CA have been produced such that BA= AD and CA=AE and given to prove  DE|| BC

Consider triangle BAC and DAE,

We have

BA = ADand CA = AE  [∴ given in the data]

And also BAC =DAE  [∴ vertically opposite angles]

So, by SAS congruence criterion, we have ΔBAC  ΔDAE

⇒ BC = DE and DEA=BCA,EDA=CBA

[Corresponding parts of congruent triangles are equal]

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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