in a quadrilateral ABCD; prove that the sum of three sides is greater than the fourth side
Asked by | 14th Oct, 2013, 01:49: PM
Join any one of the diagonal, say BD.
We know that, in a triangle, sum of two sides is greater than the third side.
Now, in triangle ABD, we have:
AB + AD > BD ... (1)
In triangle BDC, we have:
BD + CD > BC ... (2)
Adding (1) and (2), we get,
AB + AD + BD + CD > BD + BC
AB + AD + CD > BC (Subtracting BD from both sides)
Thus, we have proved that sum of three sides is greater than the fourth side.
Answered by | 14th Oct, 2013, 02:13: PM
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