Let ABC be a triangle and P be an interior point, prove that AB + BC +

1.	Let ABC be a triangle and P be an interior point, prove that AB + BC + CA < 2(PA + PB + PC).
Answer» In a triangle, sum of any two side is grater than the third side \(\therefore\) In ΔPBA, AB < PA + PB ….(i) In ΔPBC, BC < PB + PC ….(ii) In ΔPCA, AC < PC + PA ….(iii) By adding (i), (ii) and (iii) AB + BC + AC < PA + PB + PB + PC + PC + PA AB + BC + AC < 2 (PA+ PB + PC).

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

Answer»

In a triangle, sum of any two side is grater than the third side

\(\therefore\) In ΔPBA, AB < PA + PB ….(i)

In ΔPBC, BC < PB + PC ….(ii)

In ΔPCA, AC < PC + PA ….(iii)

By adding (i), (ii) and (iii)

AB + BC + AC < PA + PB + PB + PC + PC + PA

AB + BC + AC < 2 (PA+ PB + PC).