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.

Solution:

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).

Proof:

Statement	Reason
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.

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

[?]Subscribe To This Site

E-mail Address

First Name

Then Don't worry — your e-mail address is totally secure. I promise to use it only to send you Math Only Math.