Question

In the given figure, △ABC  and Prove that . ar(△ABC) ar(△DBC) = AO DO △DBC​

Answer 1

Given :

➤ ∆ABC and ∆DBC are on the same base DC and AD intersect BC at O.

To Prove :

➤ $\sf{\dfrac{ar(\triangle ABC)}{ar(\triangle DBC)} = \dfrac{AO}{DO}}$

Construction :

➤ Draw AL ⟂ BC and DM ⟂ BC.

Proof :

In ∆ALO and ∆DMO, we have

⪼ ∠ALO - ∠DMO - 90°

⪼ ∠ALO = ∠DMO [vertical opposite angle]

∴ ∆ALO ~ ∆DMO [bf AA similarity]

$\sf{: \implies \dfrac{AL}{DM} = \dfrac{AO}{DO}\:\:-----1}$

$\sf{: \implies \therefore \dfrac{ ar(\triangle ABC)}{ar(\triangle DBC)}}$

$\sf{: \implies \dfrac{ \dfrac{1}{2} \times BC \times AL }{ \dfrac{1}{2} \times BC \times DM} = \dfrac{AL}{DM} = \dfrac{AO}{DO} \: \: \bigg[ using \: equation \: 1\bigg]}$

$\bf{Hence, \dfrac{ar(\triangle ABC)}{ar(\triangle DBC)} = \dfrac{AO}{DO}}$

Hence, Proved !!