Q) In the given figure, Δ ABC and ΔDBC are on the same base BC. If AD intersects BC at O, prove that \frac{ar(\triangle ABC)} {ar(\triangle DBC)} = \frac{AO} {DO}

In the given figure, Δ ABC and ΔDBC are on the same base BC  10th Board CBSE

Ans: Let’s draw perpendicular from points A and D on line BC:

In the given figure, Δ ABC and ΔDBC 10th Board CBSE

In Δ AON and Δ DOM,

∠ AON = ∠ DOM (interior angles)

∠ ANO = ∠ DMO (given that AN and DM are \perp to BC)

\therefore  Δ AON \sim Δ DOM

Hence, \frac{AN}{DM} = \frac{AO}{DO}………….. (i)

Now let’s calculate, \frac{ar(\triangle ABC)} {ar(\triangle DBC)}:

= \frac{\frac{1}{2} (BC)(AN)} {\frac{1}{2} (BC) (DM)}

= \frac{AN} {DM}

by substituting values from equation (i), we get:

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

Hence proved.

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