Q. 7

The tangent to the circumcircle of an isosceles ΔABC at A, in which AB = AC, is parallel to BC.

Answer :

True

Let us consider a circle in which EF is a tangent passing through point A on the circle and ABC is an isosceles triangle in the circle , in which AB = AC



To Prove : EF || BC


Construction : Join OA , OB and OC


Proof :


AB = AC [Given]


ACB = ABC [Angles opposite to equal sides are equal] [1]


EAB = ACB [Tangents drawn from an external point to a circle are equal] [2]


From [1] and [2]


EAB = ACB


i.e. EF || BC [ two lines are parallel if their alternate interior angles are equal]


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