Sum of the Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals

Here we will prove that in any quadrilateral the sum of the four sides exceeds the sum of the diagonals.


Given: ABCD is a quadrilateral; AC and BD are its diagonals.

Sum of the Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals

To prove: (AB + BC + CD + DA) > (AC + BD).




1. In ∆ADB, (DA + AB) > BD.

1. Sum of the two sides of a triangle is greater than the third side.

2. In ∆ABC, (AB + BC) > AC.

2. As above.

3. In ∆BCD, (BC + CD) > BD.

3. As above.

4. In ∆CDA, (CD + DA) > AC.

4. As above.

5. 2(AB + BC + CD + DA) > 2(AC + BD).

5. Adding the equations in statements 1, 2, 3 and 4.

6. (AB + BC + CD + DA) > (AC + BD). (Proved)

6. Cancelling the common factor 2.

9th Grade Math

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