Given: AB = BC, AC is ∠ bisector of ∠BAD Prove: BC ∥ AD

<BAC ≅ BCA by rule Base angle theorem

Step-by-step explanation:

What we know:

BAC = DAC

BC = BA

ΔBCA is an isosceles so ∠BCA  =  ∠DCA and ∠BAC

and we found that out by base angle theorem since

Base angle Theorem  = Two base angles of a icosceles triangle are equal.

And since ΔBCA is an isoceles then ∠A and ∠C will be equal. And so we can prove BC is a parallel to AD

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