In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

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Question 3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that $\dfrac{ar (ABC)}{ar (DBC)} = \dfrac {AO}{DO}$

Solution:

Given,

ABC and DBC are two triangles on the same base BC. AD intersects BC at O

We have to prove:

$\dfrac{Area (ΔABC)}{Area (ΔDBC)}$ = $\dfrac{AO}{DO}$

Let us draw two perpendiculars AP and DM on line BC

We know that area of a triangle = $\dfrac{1}{2}$ × Base × Height

$\dfrac{ar (\triangle ABC)}{ar(\triangle DEF)} = \dfrac{\frac{1}{2}\times BC \times AP}{\frac{1}{2}\times BC \times DM} = \dfrac{AP}{DM}$

In ΔAPO and ΔDMO

$\angle APO = \angle DMO$ (Each 90°)

$\angle AOP = \angle DOM$ (Vertically opposite angles)

∴ ΔAPO ~ ΔDMO (AA similarity criterion)

∴$\dfrac{AP}{DM} = \dfrac{AO}{DO}$

$\dfrac{Area (ΔABC)}{Area (ΔDBC)} = \dfrac{AO}{DO}$

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