In a quad. Abcd ,prove that ab+bc+cd>ad
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Draw AC in the quadrangle ABCD.
In ΔACD, we have AD - CD < AC
in triangle ABC, we have AC < AB + BC
Hence, by combining the above two: AD - CD < AC < AB + BC
So AB + BC > AD - CD
or AB + BC + CD > AD
In ΔACD, we have AD - CD < AC
in triangle ABC, we have AC < AB + BC
Hence, by combining the above two: AD - CD < AC < AB + BC
So AB + BC > AD - CD
or AB + BC + CD > AD
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