In Fig. 5, $ABC$ and $DBC$ are two triangles on the same base $BC$. If $AD$ intersects $BC$ at $O$, show that
$$\frac{ ext{ar}( riangle ABC)}{ ext{ar}( riangle DBC)} = \frac{AO}{DO}$$

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Question

In Fig. 5, $ABC$ and $DBC$ are two triangles on the same base $BC$. If $AD$ intersects $BC$ at $O$, show that
$$\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle DBC)} = \frac{AO}{DO}$$

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Accepted Answer

Step 1: Draw $AX \perp BC$ and $DY \perp BC$. $	riangle AOX \sim 	riangle DOY$ (AA similarity)
Step 2: $\frac{AX}{DY} = \frac{AO}{DO}$ ... (i)
Step 3: $\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle DBC)} = \frac{\frac{1}{2} 	imes BC 	imes AX}{\frac{1}{2} 	imes BC 	imes DY} = \frac{AX}{DY}$
Step 4: $\frac{AX}{DY} = \frac{AO}{DO}$ (From (i)). Hence $\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle DBC)} = \frac{AO}{DO}$

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