Question

if triangle ABC is isosceles with ab=acb prove that the tangent at A to the circumcircle of triangle abc is parallal to bc

Answer 1

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.

Answer 2

Let DAE be tangent at A to the circumcircle of △ABC.

In △ABC,

⇒ AB=AC [ Given ]

∴ ∠ACB=∠ABC ----- ( 1 ) [ Angles opposite too equal sides are equal ]

According to alternate segment theorem, the angle between the tangent and chord at the point of contact is equal to the angles made by the chord in the corresponding alternative segment.

DAE is the tangent and AB is the chord.

∴ ∠DAB=∠ACB ----- ( 2 )

From ( 1 ) and ( 2 ),

⇒ ∠ABC=∠DAB

If a transversal intersects two lines such that a pair of alternate interior angles are equal, then two lines are parallel.

∴ DE∥BC

∴ The tangent at A to the circumcircle of △ABC is parallel BC.

______________________

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