Sarthaks Test
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O is any point in the interior of ΔABC. Prove that

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + QB + OC

(iii) OA + OB + OC >1/2 x (AB + BC + CA)

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Given that O is any point in the interior of ΔABC.

We have to prove

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + QB + OC

(iii) OA + OB + OC >1/2 x (AB + BC + CA)

We know that, in a triangle the sum of any two sides is greater than the third side

So, we have

Now, extend (or) produce BO to meet AC in D.

Now, in ΔABD,we have

Similarly in ΔODC, we have

OD+DC>OC   ....(5)

(i) Adding (4) and (5), we get

(ii) Adding equation (6), (7) and (8), we get

(iii) Adding equations (1), (2) and (3)

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