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In a quadrilateral ABCD, (A B+B C+C D+D A)<(A C+B D)

Option: 1

True


Option: 2

False


Answers (1)

best_answer

\\\text{Join A C . Then}, \\ A B+B C>A C \text { and } C D+D A>A C \\ \therefore A B+B C+C D+D A)>2 A C \\ \text{Similarly, by joining BD, we get} \\ (A B+B C+C D+D A)>2 B D \\ \text{Adding respective sides of (i) and (ii), we get the required result.}

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seema garhwal

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