Question

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

Answer 1

E is the midpoint of CA.
So, AE = EC .....(1)
Also, BD = $\frac{1}{2}$ CA (Given)
So, BD = AE .....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
BE acts as the median of $△$ABC
so, ar($△$BCE) = ar($△$ABE) = $\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$ .....(1)
ar($△$DBC) = ar($△$BCE) .....(2) (Triangles on the same base and between the same parallels are equal in area)
From (1) and (2)
ar(∆ABC) = 2ar(∆DBC)