Question

In Fig., BD || CA, E is mid-point of CA and BD = 1/2 CA. Prove that ar (ABC) = 2ar (DBC)

Answer 1

Join DE. Here BCED is a parallelogram, since 

BD = CE and BD || CE 

ar (DBC) = ar (EBC) (1) [Have the same base BC and between the same parallels] 

In ∆ ABC, BE is the median, 

So, ar (EBC) = 1/2 ar (ABC) 

Now, ar (ABC) = ar (EBC) + ar (ABE) 

Also, ar (ABC) = 2 ar (EBC), therefore, 

ar (ABC) = 2 ar (DBC).