Question

In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that ar(∆ABC)/ar(∆DCB) = AO/DO.

Answer 1

construction ∶ draw AX⊥CO and DY⊥BO. 

As,

In ∆ ABC and ∆DBC, ∠AXY = ∠DYO = 90° (By constructin)

∠AOX = ∠DOY (vertically opposite angle) 

∴ ∆AXO~∆DYO(By AA criterion) 

∴ AX/DY = AO/DO (Thales's theorem) … (ii)From (i)and (ii), we have ∶ ar(∆ABC)/ar(∆DBC)

This completes the proof.