Question

In the given figure, BD || CA, E is the midpoint of CA and $BD=\frac{1}{2}CA$. Prove that ar(∆ABC) = 2 × ar(∆DBC).

Answer 1

Given: BD || CA and E is the midpoint of CA.
To prove: ar(∆ABC) = 2 × ar(∆DBC)​
Construction: Join DE.
Proof:
Now, BD || CE and BD = CE [ E is the mid point of AC]
∴ BCED is a parallelogram.

So, ar(∆ EBC) = ar(∆DBC) ...(i) [On the same base and between the same parallel lines]
∵ ar(∆EBC) = $\frac{1}{2}$× ar(∆ABC) ...(ii) [ BE is the median of ∆ABC]
From equation (i) and (ii), we get:
ar(∆DBC) = $\frac{1}{2}$ × ar(∆ABC)
⇒ ar(∆ABC) = 2 ×​ ar(∆DBC)
Hence, proved.