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Question
The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.
Options
True
False
Solution
This statement is True.
Explanation:
Let EAF be tangent to the circumcircle of ∆ABC.
To prove: EAF ॥ BC
We have, ∠EAB = ∠ACB ...(i) [Angle between tangent and chord is equal to angle made by chord in the alternate segment]
Here, AB = AC
⇒ ∠ABC = ∠ACB ...(ii)
From equation (i) and (ii), we get
∠EAB = ∠ABC
∵ Alternate angles are equal.
⇒ EAF ॥ BC
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