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Standard IX

Mathematics

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

If O is any p...

Question

If O is any point within $△$ ABC,then to prove that AB + BC + CA > AO + BO + CO

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Solution

Let us $△$ ABC and mark point O anywhere inside the triangle.

Now, there is a known inequality:
AO + CO < AB + BC (Proof can be given, if you need)
Similarly,
AO + BO < AC + BC
BO + CO > AB + AC
Adding all these together, we get,
2(AO + BO + CO) < 2 (AB + BC + CA)
$\to A B + B C + C A > A O + B O + C O$

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If O is any point within △ABC,then to prove that AB + BC + CA > AO + BO + CO

Let us △ABC and mark point O anywhere inside the triangle. Now, there is a known inequality: AO + CO < AB + BC (Proof can be given, if you need) Similarly, AO + BO < AC + BC BO + CO > AB + AC Adding all these together, we get, 2(AO + BO + CO) < 2 (AB + BC + CA) →AB+BC+CA>AO+BO+CO

If O is any point within $△$ ABC,then to prove that AB + BC + CA > AO + BO + CO