Question

a quadrilateral ABCD is drawn to circumscribe a circle prove that a
b + c d is equal to AD + BC

Answer 1

Proof - Let AB touches the circle at P. BC touches the circle at Q. DC touches the circle at R. AD touches the circle at S.

THEN,PB=QB  ( Length of the the tangents drawn from the external point are always equal )

QC =RC " 

AP=AS  "

DS=DP  "

NOW,  AB + CD

= AP + PB+DR+RC

= AS+QB+DS+CQ

= AS+DS+QB+CQ

= AD+BC

"HENCE PROVED"

Answer 2

Given :-

A quadrilateral ABCD is drawn to circumscribe a circle.

To prove :-

AB + CD = AD + BC

Proof :-

The lengths of tangents drawn from an external point to a circle are equal.

=>B is any point outside the circle and BP , BQ are tangents to the circle.

Now ,

BP = BQ ... (1)

AP = AS ... (2)

CR = CQ ...(3)

DR = DS ...(4)

Adding (1), (2), (3) and (4), we get (BP+AP) + (CR+DR) = (BQ+CQ) +(AS+DS)

AB + CD = BC + AD

Hence proved

Additional Information :-

Theorem : The tangent at any point of a circle is perpendicular to the radius through the point of contact .

Theorem : The lengths of tangents drawn from an external point to a circle are equal .