Question

A quadrilateral ABCD is drawn to circumscribe a circle as shown in the figure. Prove that AB + CD = AD + BC​

Answer 1

Step-by-step explanation:

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Answer 2

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Given - c(o,r)

To Prove - AB + CD = AD + BC

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