proof that ab+ac+BC>2ab
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Step-by-step explanation:
We know that in a triangle Sum of Two Sides > Third Side
∴ In ΔABD
AB + BD > AD --- ( i )
In ΔACD
AC + CD > AD ----- ( ii )
Adding equation ( i ) and equation ( ii )
AB + BD + AC + CD > AD + AD
AB + BC + AC > 2AD [ ∵ BD + CD = BC]
Hence Proved
SohelMJ:
thanks for the answer
pleasure :)
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