If four sides of a quadrilateral $A B C D$ are tangential to a circle, then

A C + A D = B D + C D

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A B + C D = B C + A D

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A B + C D = A C + B C

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A C + A D = B C + D B

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Solution

The correct option is B $A B + C D = B C + A D$
A circle has been inscribed in a quadrilateral $A B C D$ .
To find out- which of the options is true.
Solution :
Let the circle touch the side $A B$ at $P$ , $B C$ at $Q$ , $C D$ at $R$ and $A D$ at $S$ .
$∴$ $A P = A S, B P = B Q, C R = C Q$ and $D R = D S$ , since the lengths of the tangents, drawn from a point to a circle, are equal.
$∴ A P + B P = A B = A S + B Q$ .....(i)
$C R + D R = C D = C Q + D S$ ......(ii).
Adding (i) and (ii), we get
$A B + C D = A S + B Q + C Q + D S = (A S + D S) + (B Q + C Q) = A D + B C$
Ans : Option B

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