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Question
In the given figure, prove that:
CD + DA + AB + BC > 2AC
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Solution
We have to prove that CD + DA + AB + BC > 2AC
In ΔABC we have
AB + BC > AC (As sum of two sides of triangle is greater than third one) ........(1)
In ΔACDwe have
AD + CD > AC (As sum of two sides of triangle is greater than third one) .........(2)
Hence
Adding (1) & (2) we get AB + BC + AC + CD > 2AC Proved.
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