In the figure below, $ \Delta A B C $ and $ \Delta D B C $ are on the same base $ B C $. If $ A D $ and $ B C $ intersect at $O$, prove that $ \frac{\text { Area }(\Delta A B C)}{\text { Area }(\Delta D B C)}=\frac{A O}{D O} $
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Given:

\( \Delta A B C \) and \( \Delta D B C \) are on the same base \( B C \).

\( A D \) and \( B C \) intersect at $O$.
To do:
We have to prove that  \( \frac{\text { Area }(\Delta A B C)}{\text { Area }(\Delta D B C)}=\frac{A O}{D O} \).

Solution:

We know that,

Area of triangle $=\frac{1}{2}\times base \times height$

Therefore,

$\frac{ar(\triangle ABC)}{ar(\triangle DBC)}=\frac{\frac{1}{2}\times BC \times AL}{\frac{1}{2}\times BC \times DM}$

$=\frac{AL}{DM}$.....(i)

In $\triangle ALO$ and $\triangle CMO$,

$\angle LOA=\angle MOC$      (Vertically opposite angles)

$\angle ALO=\angle CMO=90^o$

Therefore,

$\triangle ALO \sim\ \triangle CMO$   (By AA similarity)

This implies,

$\frac{AL}{DM}=\frac{AO}{DO}$....(ii)    (Corresponding parts of similar triangle are proportional)

From equations (i) and (ii), we get,

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

Hence proved.