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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=ACABC=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=ABCEAFBC
( Since, the alternate interior angles are equal)


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