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

# 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.

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)

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.

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)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...