Question

In the given figure,$\triangle ABC and \triangle DBC$ are on the same base BC. If AD and BC intersect at O, prove that Area $\frac{Area(\triangle ABC)}{Area(\triangle DBC)} = \frac{AO}{DO}$

Answer 1

SOLUTION :  

Given :  

ΔABC and ΔDBC are on the same BC. AD and BC intersect at O.

Prove that : ar(ΔABC) / ar(ΔDBC) = AO/DO

Draw AL⊥BC and DM⊥BC

In ΔALO and ΔDMO, we have

∠ALO = ∠DMO        [each 90°]

∠AOL =∠DOM   [vertically opposite angles]

Therefore,  ΔALO∼ΔDMO

[By AA similarity criterion]

∴ AL/ DM = AO/DO ………….(1)

[Corresponding sides of two similar triangles are proportional]

arΔABC/arΔBCD = (½ × BC × AL)/ (½ × BC × DM)

[Area of ∆ = ½ × base × height]

arΔABC/arΔBCD = AL/DM

arΔABC/arΔBCD = AO/DO

[From eq 1]

Hence, proved arΔABC/arΔBCD = AO/DO

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