In fig 6.44 abc and dbc are two triangles

Asked by admin @ 16/08/2021 in Math viewed by 368 People

ABC and DBC are two triangle on the same base BC .if AD intersect BC at O show that

ar(ABC)/ar(DBC)=AO/DO

Answered by admin @ 16/08/2021

Given:

ABC & DBC are two triangles on the same base BC

AD intersects BC at O

To show:

$\frac{Ar (\triangle ABC)}{Ar (\triangle DBC)} = \frac{AO}{DO}$

Solution:

Construction:- AP ⊥ BC and DQ ⊥ BC

In Δ APO and Δ DQO, we have

∠APO = ∠DQO = 90° .... [right angles]

∠AOP = ∠DOQ ..... [vertically opposite angles]

∴ Δ APO ~ Δ DQO ...... [By AA similarity]

We know that → the corresponding sides of similar triangles are proportional to each other.

∴ $\frac{AP}{DQ} = \frac{AO}{DO}$ ...... (i)

Now, we have

$\frac{Area (\triangle ABC)}{Area (\triangle DBC)} = \frac{\frac{1}{2}\times AP\times BC}{\frac{1}{2}\times DQ\times BC}$

cancelling the similar terms

$\implies \frac{Area (\triangle ABC)}{Area (\triangle DBC)} = \frac{ AP}{ DQ}$

from (i), we get

$\implies \frac{Area (\triangle ABC)}{Area (\triangle DBC)} = \frac{ AP}{ DQ} = \frac{AO}{DO}$

$\implies \boxed{\bold{\frac{Area (\triangle ABC)}{Area (\triangle DBC)} = \frac{AO}{DO}}}$

Hence proved

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