Question

In the figure, O is the interior point of
AABC. BO meets AC at D. Show that OB + OC < AB + AC.​

Answer 1

In ∆ABD, AB+AD>BD ....(i)

We know that the sum of any two sides of a triangle is greater than the third side. Also, we have

BD=BO+OD

AB+AD>BO+OD ....(ii)

Similarly, in ∆ COD, we have

OD+DC>OC .....(iii)

On adding (ii) and (iii), we have

AB+AD+OD+DC>BO+OD+OC

=> AB+AD+DC>BO+OC

=> AB+AC>BO+OC

or BO+OC<AB+AC

Hence, proved.