Question

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

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Answer 1

Answer:

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Answer 2

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$\purple{\bigstar}$ $\text{\Large\underline{\bf{Question:-}}}$

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

$\blue{\bigstar}$ $\text{\Large\underline{\bf{Given:-}}}$

Given c(o,r) proof-

let AB touches the circle at P, BC at Q, DC at R and AD at S.

then PB= PQB(length of tangents drawn from an 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

$\large\bf\star\pink{\underline{\overline{ \: hence \: proved \: }}}\star$

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