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

Given: Δ ABC and Δ DBC are on the same base BC

_{To prove:}

Theorem Used:

If two triangles are similar, then the ratio of their corresponding sides are equal.

Explanation:

_{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 ΔALO and ΔDMO,}

_{∠ALO = ∠DMO (Each is 90°)}

_{∠AOL = ∠DOM (Vertically opposite angle)}

_{∴ ΔAPO ~ ΔDMO (By AA rule)}

Hence Proved