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Prove that the sum of the two diagonals of a quadrilateral is less than its perimeter

General enunciation: We have to prove that the sum of the two diagonals of a quadrilateral is less than its perimeter.

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Particular enunciation: Let, AC and BD are the two diagonals od ABCD quadrilateral.

Prove that AC + BD < AB + BC + CD + AD

Proof: ABC, AB + BC > AC ——————–(i)

BCD, BC + CD > BD —————————–(ii)

ADC, AD + CD > AC —————————–(iii)

ABD, AB + AD > BD ——————————(iv)

By adding (i), (ii), (iii) and (iv) we get

AB + BC + BC + CD + AD + CD + AB + AD > AC + BD + AC + BD

⟹ 2(AB + CD + BC + BD) > 2(AC + BD)

⟹ AB + CD + BC + BD > AC + BD

Therefore, AC + BD < AB + CD + BC + BD. (Proved)

Problem: The diagonals AC and BD of a quadrilateral ABCD intersect at O. Prove that AC + BD > BC + AD.

General enunciation: We have to prove that the diagonals AC and BD of a quadrilateral ABCD intersect at O. Prove that AC + BD > BC + AD.

Screenshot_61

 

Particular enunciation: Let, the diagonal AC and BD of a quadrilateral ABCD intersect at O. Prove that AC + BD > BC + AD.

Proof: In AOD, AO + OD > AD —————–(i)

In BOC, BO + CO > BC —————————(ii)

By adding (i) and (ii) we get,

AO + DO + BO + CO > AD + BC

⟹ (AO + CO) + (DO + BO) > AD + BC

⟹ AC + BD > AD + BC

Therefore, AC + BD > BC + AD. (Proved)

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