Question

When two parallel tangents drawn of a circle to meet a third tangent, how do you prove that the sum of interior angles of a transversal is equal to 180?​

Answer 1

Two parallel tangents of a circle and third tangent intersects at P & Q.

AQBP is a quadrilateral.

⇒ ∠A + ∠B = 90 + 90 = 180

{Radius through point of contact is perpendicular to tangent}

⇒ ∠A + ∠B + ∠P + ∠Q = 360°

⇒ 180° + ∠P + ∠Q = 360°

⇒ ∠P + ∠Q = 180°

⇒ ∠APO = ∠OPQ = (1/2) ∠P

∠BQO = ∠PQO = (1/2) ∠Q

  2∠OPQ + 2∠PQO = 180°

Answer 2

Answer:

Two parallel tangents of a circle and third tangent intersects at P & Q.

AQBP is a quadrilateral.

⇒ ∠A + ∠B = 90 + 90 = 180

{Radius through point of contact is perpendicular to tangent}

We know that in the quadrilateral, the sum of all the four corners is 360.

⇒ ∠A + ∠B + ∠P + ∠Q = 360°

⇒ 180° + ∠P + ∠Q = 360°

⇒ ∠P + ∠Q = 180°

⇒ ∠APO = ∠OPQ = (1/2) ∠P

∠BQO = ∠PQO = (1/2) ∠Q

  2∠OPQ + 2∠PQO = 180°

HENCE PROVED