Question 7
The tangents to the circumcircle of an isosceles ΔABC at A, which AB=AC, is parallel to BC.
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Solution
True
Let EAF be tangents to the circumcircle of ∆ABC
To prove: EAF ∥ BC Here,AB=AC⇒∠ABC=∠ACB...(i) [Base angles of an isosceles triangle are equal]
Angle between a tangent and a chord is equal to the angle made by chord in the alternate segment. ∴∠EAB=∠BCA...(ii)
From Eqs. (i) and (ii), we get ∠EAB=∠ABC⇒EAF∥BC
( Since, the alternate interior angles are equal)