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. |
From Sum of the Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals to HOME PAGE
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.
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.