In figure, ABC and DBC are two triangles on the same base BC.
Question.

In figure, $\mathrm{ABC}$ and $\mathrm{DBC}$ are two triangles on the same base $\mathrm{BC}$. If $\mathrm{AD}$ intersects $\mathrm{BC}$ at $\mathrm{O}$, show that $\frac{\operatorname{ar}(A B C)}{\operatorname{ar}(D B C)}=\frac{A O}{D O}$.’

Solution:

Draw $\mathrm{AL} \perp \mathrm{BC}$ and $\mathrm{DM} \perp \mathrm{BC}$ (see figure)

$\Delta \mathrm{OLA} \sim \Delta \mathrm{OMD}$(AA similarity criterion)

$\Rightarrow \frac{A L}{D M}=\frac{A O}{D O}$ …(1)

Now, $\quad \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D B C)}=\frac{\frac{1}{2} \times(B C) \times(A L)}{\frac{1}{2} \times(B C) \times(D M)}$

$=\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}$ (by1)

Hence, $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D B C)}=\frac{A O}{D O}$
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