Question

8. In Fig. 14.172, DE is a tangent to the circumcircle of
A ABC at the vertex A such that DE || BC. Show that
AB = AC​

Answer 1

Answer:

According to Alternate Segment Theorem,

   The angle formed by a chord with a tangent drawn to the chord's any

   endpoint is equal to any angle in the alternate segment. .

Here, the tangent DE has A as its point of contact with the circle, and A is an endpoint of the chord AC.

So, by Alternate Segment Theorem,

                ∠DAB = ∠ACB                                            (I)

Now, by considering DE || BC and AB as the transversal,

                 ∠DAB = ∠ABC                                           (ii)

From (i), (ii),

                  ∠ACB = ∠ABC

We know that in a triangle sides opposite equal angles are equal

So, as in ΔABC, ∠ACB = ∠ABC,

                 AB = AC

Hence proved that AB = AC

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