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.
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. |
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