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