Question

9. In the Fig., AD is one of the medians of a AABC and P is a point on AD. A 15. Р B D C Prove that (i) ar (ABDP)= ar (ACDP) (ii) ar (AABP) = ar (AACP).​

Answer 1

Step-by-step explanation:

Given:△ABC in which AD is the median.

P is any point on AD.

Join PB and PC.

To prove:

(i) Area of ΔPBD= area of ΔPDC.(ii) Area of ΔABP= area of ΔACP. Proof: From fig (1) AD is a median of △ABC.

So, ar (ΔABD)=ar(ΔADC)….(1)

Also, PD is the median of △BPD

Similarly, ar (ΔPBD)=ar(ΔPDC)…. (2)

Now, let us subtract ( 2 ) from ( 1 ), we get ar (ΔABD)−ar(ΔPBD)=ar(ΔADC)−ar(ΔPDC)

Or ar (ΔABP)=ar(ΔACP)

Hence proved.