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In a quadrilateral ABCD, (A B+B C+C D+D A) < 2(B D+A C) 

Option: 1

True


Option: 2

False


Answers (1)

best_answer

\\O A+O B>A B, O B+O C>B C \\\\ O C+O D>C D, O D+O A>D A\\ \\\text{Adding, we get}\\\\ 2(O A+O C+O B+O D)>(A B+B C+C D+D A) \\ \\\Rightarrow \quad 2(A C+B D)>(A B+B C+C D+D A) \\ \\\because O A+O C=A C, O B+O D=B D

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Divya Prakash Singh

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