**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)

**∆ **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.

**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)**