Question

In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that ar(∆ABD) : ar(∆ADC) = m : n.

Answer 1

Given: D is a point on BC of ∆ ABC, such that BD : DC = m : n
To prove: ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC.
Proof:
ar(∆ABD)​ = $\frac{1}{2}$ ×​ BD ×​ AL ...(i)
ar(∆ADC)​ = $\frac{1}{2}$​ ×​ DC ×​ AL ...(ii)
Dividing (i) by (ii), we get:
$$\begin{gathered} \frac{\mathrm{ar}(△ABD)}{\mathrm{ar}(∆ADC}=\frac{{\displaystyle \frac{1}{2}}\times BD\times AL}{{\displaystyle \frac{1}{2}}\times DC\times AL} \\ =\frac{BD}{DC} \\ =\frac{m}{n} \end{gathered}$$

∴ ar(∆ABD) : ar(∆ADC) = m : n