InterviewSolution
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InterviewSolution
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In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that(i) AB + BC + CD + DA > 2AC(ii) AB + BC + CD > DA(iii) AB + BC + CD + DA > AC + BD. |
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Answer» (i) Consider △ ABC We know that AB + BC > AC …… (1) Consider △ ACD We know that AD + CD > AC ……. (2) By adding both the equations we get AB + BC + AD + CD > AC + AC So we get AB + BC + AD + CD > 2AC …….. (3) Therefore, it is proved that AB + BC + AD + CD > 2AC. (ii) Consider △ ABC We know that AB + BC > AC Add CD both sides of the equation AB + BC + CD > AC + CD …… (4) Consider △ ACD We know that AC + CD > DA ……. (5) By substituting (5) in (4) we get AB + BC + CD > DA …….. (6) (iii) Consider △ ABD and △ BDC We know that AB + DA > BD ……. (7) So we get BC + CD > BD ……. (8) By adding (7) and (8) we get AB + DA + BC + CD > BD + BD On further calculation AB + DA + BC + CD > 2BD ……. (9) By adding equations (9) and (3) we get AB + DA + BC + CD + AB + BC + AD + CD > 2BD + 2AC So we get 2 (AB + BC + CD + DA) > 2 (BD + AC) Dividing by 2 AB + BC + CD + DA > BD + AC Therefore, it is proved that AB + BC + CD + DA > BD + AC. |
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