`P` is the mid-point of side `A B` of a parallelogram `A B C D` . A line through `B` parallel to `P D` meets `D C` at `Q\ a n d\ A D` produced at `Rdot` Prove that: `A R=2B C` (ii) `B R=2\ B Q`

Explanation

Solution
(i) In `/_\ARB`, `P` is the mid-point of AB and PD∥BR.
∴ `D` is a mid-point of `AR` [ Converse mid point theorem ]
∴ `AR=2AD`
But `BC=AD` [ Opposite sides of parallelogram are equal ]

∴ `AR=2BC`
(ii) `ABCD` is a parallelogram.
∴ `DC∥AB
⇒DQ∥AB`
Now, in `/_\ARB`,
`D` is a mid-point of `AR` and `DQ∥AB`
∴ ` Q` is a mid point of `BR` [ Converse mid-point theorem ]
⇒ `BR=2BQ`