Question

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

Answer 1

ANSWER:
E is the midpoint of CA.
So, AE = EC
.....(1)
Also, BD = 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) = 1/2ar△ABC
.....(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)