Prove that the difference of any two sides of the triangle is less than the third side
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Cut off AD = AB, join BD
2 = 4
Exterior 1 > 4 (exterior angle is greater than each of the interior angles)
Exterior 2 > 3
1 > 3
BC > DC (side opposite greater angle is greater)
BC > AC - AD (DC = AC - AD)
BC > AC - AB (AD = AB)
BC > AC - AB < BC
Similarly, BC - AC < AB and BC - AC < AC
Hence, the difference of any two sides of a triangle is less than the third side
please use the image to understand properly...
and there is problem of symbol of angle please try to understand..
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