Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Or

A quadrilateral ABCD id drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

We have

Given: XY is a tangent to the circle at the point P.

To prove: Tangent at any point of a circle is perpendicular to the radius through the point of contact.

⇒ OP ⊥ XY

Construction: Take a point Q on XY other than P and then join OQ

Proof: If this point Q lies inside the circle, then XY will become a secant and will not remain a tangent to the circle anymore.

∴, OQ > OP

Similarly, this happens with every point on the line XY except the point P.

⇒ OP is the shortest of all the distances of the point O to the points of XY.

⇒ OP ⊥ XY [As the shortest of all distances from the tangent to the radius is perpendicular to the radius]

Hence, proved.

Or

We have

Given: ABCD is a quadrilateral and it circumscribe a circle that touches the quadrilateral at P, Q, R, S.

To Prove: AB + CD = AD + BC

Proof: Since tangents from an external point to the circle are equal, we can write

AP = AS …(i)

BP = BQ …(ii)

DR = DS …(iii)

CR = CQ …(iv)

Adding (i), (ii), (iii) and (iv), we get

AP + BP + DR + CR = AS + BQ + DS + CQ

⇒ (AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Hence, proved.