Question

A quadrilateral ABCD is drawn to circumscribe a circle.

Prove that AB + CD = AD + BC​

Answer 1

Step-by-step explanation:

its critical shaped drawn by 5 and perimeter this so easy a b + CD is equal area + BC inform that what is the formula of a b + CD and his people adbc is easy to BC + BC and multiply CD CD is equal to AD multiple BC

Answer 2

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$\sf\underbrace{Appropriate\:Question: }$

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

$\sf\underbrace{Required\:Answer: }$

$\bf\underline{Given: }$

• Let ABCD be the quadrilateral circumscribing the circle with centre O. The quadrilateral touches the circle at point P,Q,R and S.

$\bf\underline{To\:prove:}$

• $\sf{AB\:+\:CD\:=\:AD\:+\:BD}$

$\bf\underline{Solution:}$

from theorem 10.2, lengths of tangents drawn from external point are equal.

ㅤㅤ$\sf{⟹\:Hence,\:AP\:=\:AS\: .......(1)}$

ㅤㅤ$\sf{⟹\:BP\:=\:BQ\: .......(2)}$

ㅤㅤ$\sf{⟹\:CR\:=\:CQ\: ......(3)}$

ㅤㅤ$\sf{⟹\:DR\:=\:DS\: ........(4)}$

$\bf\underline{Adding\:(1)\:+\:(2)\:+\:(3)\:+\:(4)}$

$\sf{AP\:+\:BP\:+\:CR\:+\:DR\:=\:AS\:=\:BQ\:+\:CQ\:+\:DS}$

$\sf{(AP\:+\:BP)\:+\:(CR\:+\:DR)\:=\:(AS\:+\:DS)\:+\:(BQ\:+\:CQ)}$

$\sf{AB\:+\:CD\:=\:AD\:+\:BC}$

$\bf\underline\purple{Hence\:Proved}$

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