AD is the median of a ΔABC, prove that AB + BC + CA > 2AD
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Step-by-step explanation:
In ΔABD,
AB + BD > AD …(i)
(Sum of two sides of a triangle is greater than the third side)
Similarly, In ΔADC, we have
AC + DC > AD …(ii)
Adding (i) and (ii), we have
AB + BD + AC + DC > 2AD
⇒ AB + (BD + DC) + AC > 2AD
⇒ AB + BC + AC > 2AD
Hence, proved.
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#hear is you're answer
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