Question

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In Figure ABC and DBC are two Triangles on the same base BC if a d intersect BC at O show that :-
ar∆[ABC]\ar∆[DBC]=AO\DO


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Answer 1

$\underline{\underline{\bold{Question:}}}$

In Figure ABC and DBC are two Triangles on the same base BC if AD intersect BC at O show that :-

$\bold{\dfrac{ar\Delta ABC}{ar\Delta DBC}=\dfrac{AO}{DO}.}$

$\tt{Solution:}$

Given :

Both triangles have a common base that is BC.

To prove :

$\bold{\dfrac{ar\Delta ABC}{ar\Delta DBC}=\dfrac{AO}{DO}.}$

To construct :

In Δ ABC

• Draw AE ⊥ BC.

In ΔDBC

• Draw DF ⊥ BC.

Proof :

In Δ AOE and ΔDOF

• ∠AOE = ∠DOF = Vertically opposite angle.
• ∠AEO = ∠DFO = 90°.

So ΔAOE ~ ΔDOF (AAA criteria)

$\bold {\dfrac{AO}{DO}=\dfrac{OE}{OF}=\dfrac{AE}{DF}}$

$\boxed{\bold{Area\:of\:triangle=\dfrac{1}{2}*Base*Height}}$

So

Area of triangle ABC :

$\bold{=\dfrac{1}{2}*BC*AE.}$

Area of triangle DBC :

$\bold{=\dfrac{1}{2}*BC*DF.}$

Now

$\bold{\dfrac{ar\Delta ABC}{ar\Delta DBC}=\dfrac{\dfrac{1}{2}*BC*AE}{\dfrac{1}{2}*BC*DF}}\\\\\\\bold{=\dfrac{AE}{DF}=\dfrac{AO}{DO}}.\\\\\\\bold{Proved.}\\\\\\\boxed{\boxed{\bold{\dfrac{ar\Delta ABC}{ar\Delta DBC}=\dfrac{AO}{DO}.}}}$