In quadrilateral ABCD prove AB+BC+CD+DA is greater than 2AC
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Sum of two sides in a triangle is greater than the third side
Apply it in two triangles formed by diagonal AC and add those two inequalities
BTW-
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Apply it in two triangles formed by diagonal AC and add those two inequalities
BTW-
Mark as brainliest if u like ;)
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Answer:
In ∆ ABC
Ab + bc > ac -- 1 (sum of two sides is greater than the third side.)
In ∆ adc
Cd + Da > ac --2 (sum of two sides is greater than the third side)
On adding 1 and 2 equation, we get
Cd + Da + ab + bc > 2ac
Hope it helps
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