# The base BC of ∆ABC is divided at D such that BD

Question:

The base $B C$ of $\triangle A B C$ is divided at $D$ such that $B D=\frac{1}{2} D C$. Prove that $\operatorname{ar}(\Delta A B D)=\frac{1}{3} \times \operatorname{ar}(\Delta A B C)$.

Solution:

Given: $\mathrm{D}$ is a point on $\mathrm{BC}$ of $\triangle A B C$, such that $\mathrm{BD}=\frac{1}{2} D C$

To prove: $\operatorname{ar}(\Delta A B D)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$

Construction: Draw AL ⊥ BC.
Proof:
In ∆ABCwe have:
BC = BD + DC

⇒​ BD​ + 2 BD = 3 × BD
Now, we have:

$\operatorname{ar}(\Delta A B D)=\frac{1}{2} \times B D \times A L$

$\operatorname{ar}(\Delta A B C)=\frac{1}{2} \times B C \times A L$

$\Rightarrow \operatorname{ar}(\Delta A B C)=\frac{1}{2} \times 3 B D \times A L=3 \times\left(\frac{1}{2} \times B D \times A L\right)$

$\Rightarrow \operatorname{ar}(\Delta A B C)=3 \times \operatorname{ar}(\Delta A B D)$

$\therefore \operatorname{ar}(\Delta A B D)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$