Question

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

Answer 1

Answer:

Step-by-step explanation:to prove:

AB+CD=AD+BC

Proof :

AP=AS

PB=BQ

CQ=CR

RD=DS (the length of tangent drawn from an external point to a circle are equal)

AP+PB+RD+CR=AS+BQ+DS+CQ

=>AB+CD=AD+BC

Hence proved

Answer 2

Given:

A quadrilateral ABCD is circumscribed in a circle.

To Find:

AB + CD = AD + BC

Solution:

Firstly connecting all the points of the circle to the quadrilateral

The length of the tangents drawn from an external point to a circle is equal.

DR = DS ..(i)

BP = BQ ..(ii)

AP = AS ..(iii)

CR = CQ ..(iv)

Since they are tangents on the circle from points D, B, A, and C respectively.

Using (i), (ii), (iii), (iv) add the LHS and RHS of the above equations separately.

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

Now, we rearrange the terms,

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

Then simplifying the terms

CD + AB = BC + AD

AB+CD=AD+BC

Hence proved, if a quadrilateral ABCD is drawn to circumscribe a circle then AB+CD=AD+BC.