Question

The base BC of ∆ABC is divided at D such that $BD=\frac{1}{2}DC$. Prove that $\mathrm{ar}(∆ABD)=\frac{1}{3}\times \mathrm{ar}(∆ABC)$.

Answer 1

Given: D is a point on BC of ∆ABC, such that BD = $\frac{1}{2}$DC
To prove: ar(∆ABD) = $\frac{1}{3}$ar(∆ABC)
Construction: Draw AL ⊥ BC.
Proof:
In ∆ABC, we have:
BC = BD + DC
⇒​ BD​ + 2 BD = 3 × BD
Now, we have:
ar(∆ABD)​ = $\frac{1}{2}$​ ×​ BD ×​ AL
ar(∆ABC)​ = $\frac{1}{2}$​ ×​ BC ×​ AL
⇒ ar(∆ABC) = $\frac{1}{2}$ ×​ 3BD ×​ AL = 3 ×​ $\left(\frac{1}{2}\times BD\times AL\right)$
⇒ ar(∆ABC)​ = 3 × ar(∆ABD)
∴ ​ar(∆ABD) = ​$\frac{1}{3}$​ar(∆ABC)