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If the sides of a quadrilateral touch a circle, prove that the sum of a pair of opposite sides is equal to the sum of the other pair.
Given:
The sides of a quadrilateral touch a circle.
To do:
We have to prove that the sum of a pair of opposite sides is equal to the sum of the other pair.
Solution:
Let the sides of a quadrilateral $PQRS$ touch the circle at $A, B, C$ and $D$ respectively.
Proof:
$PA$ and $PD$ are the tangents to the circle from $P$.
This implies,
$PA = PD$..….(i)
Similarly,
$QA = QB$……(ii)
$RC = RB$.….(iii)
$SC = SD$.….(iv)
Adding equations (i), (ii), (iii) and (iv), we get,
$PA + QA + SC + RC = RB + QB + SD + PD$
$PQ + RS = RQ + PS$
Hence proved.
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