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:
Area of a triangle = $\frac{1}{2}$ ×​ base ×​ height
ar(∆ABD)​ = $\frac{1}{2}$ ×​ BD ×​ AL ...(i)
ar (∆ ADC)​ = $\frac{1}{2}$​ ×​ DC ×​ AL ...(ii)
Dividing equation (i) by (ii), we get:
$$\begin{gathered} \frac{\mathrm{ar}(∆ABD)}{\mathrm{ar}(∆ADC)}=\frac{{\displaystyle \frac{1}{2}}\times BD\times AL}{{\displaystyle \frac{1}{2}}\times DC\times AL} \\ =\frac{BD}{DC} \\ =\frac{m}{n}\left(\because \frac{BD}{DC}=\frac{m}{n}\right) \end{gathered}$$
∴ ar(∆ABD) : ar(∆ADC)​ = m : n