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

Explanation

Let EAF be tangent to the circumcircle of `DeltaABC`

To prove `EAF//BC`
`angleEAB=angleABC`
Here, AB=AC
`rArrangleABC=angleACB….(i)`
[angle between tangent and is chord equal to angle made by chord in the alternate segmaent]
:.Also, `angleEAB=angleBCA....(ii)`
From Eqs. (i) and (ii), we get
`angleEAB=angleABC`
`rArrEAF||BC`