if triangle ABC is isosceles with AB=AC, prove that the tangent A to the circumcircle of triangle ABC is parallel to BC.

 

 pls clear the question........ or give diagram

give diagram plss

i also dont noe

 Let DAE be tangent at A to the circumcicle of ΔABC.

In ΔABC,

AB = AC  (Given)

∴ ∠ACB = ∠ABC  ...(1)  (Equal sides have equal angles opposite to them)

We know that, If a line touches a circle and from the point of contact, a chord is drawn, then the angles between  the tangent and the chord are respectively equal to the angles in the corresponding alternative segment.

Now, DAE is the tangent and AB is the chord.

∴ ∠DAB = ∠ACB  ...(2)

From (1) and (2), we have

∠ABC = ∠DAB

∴ DE || BC  (If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel)

Thus, the tangent at A to the circumcircle of  ΔABC is parallel BC.

If a line touches a circle and from the point of contact, a chord is drawn, then the angles between the tangent and the chord are respectively equal to the angles in the corresponding alternative segment.

cant we do without this above theorem actually we dont have this theorem

other wise can u say how to prove the above theorm