In Fig. 4.179, are on the same base BC. If AD and BC intersect at O. Prove that


We know that area of a triangle = 1/2 x base x height


Since, ΔABC and ΔDBC are one same base.


Therefore ratio between their areas will be as ratio of their heights.


Let us draw two perpendiculars AP and DM on line BC


In ΔAPO and ΔDMO,


<APO = < DMO (Each is 90°)


<AOP = <DMO (Vertically opposite angle)


<OAP = <ODM (remaining angle)


Therefore ΔAPO ~ ΔDMO (By AAA rule)


Therefore AP/ DM = AO/DO


There fore area (ΔABC/areaΔDBC) = AO/DO


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