Question

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

Answer 1

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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Also View:

triangle PQR and triangle SQR are isosceles triangles on the same base . prove that angle PQS = angle PRS.​

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triangle PQR and triangle QSR lie on same base QR.Also angle PSQ=angle RQS.if ar triangle PQR=12cmsquare,find ar triangle QSR

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Answer 2

Step-by-step explanation:

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