Q5 of 49 Page 61

Let ABC be a triangle and P be an interior point. Prove that AB + BC + CA < 2 (PA + PB + PC).

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According to triangle inequality the sum of smaller two sides is always greater than the largest side


In ΔAPB


AP + PB>AB …(i)


In ΔAPC


AP + PC>AC …(ii)


In ΔCPB


CP + PB>CB …(iii)


Adding (i),(ii) and (iii) we get


2 (PA + PB + PC)> AB + BC + CA


or changing the sign of inequality we can say


AB + BC + CA < 2 (PA + PB + PC)


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