In a parallelogram `A B C D ,` the bisector of `/_A` also bisects `B C` at `Xdot` prove that `A D=2A Bdot`

Explanation

Given: In parallelogram `ABCD`, bisector of `∠A` bisects `BC` at `X`.
To Prove: `AD=2AB`
Proof: `AD` parallel to `BC` and `AX` cuts them
`=>/_DAX=/_AXB` (alternate angles)
`/_DAX=/_XAB` (`AX` is bisector of `∠A`)
`:./_AXB=/_XAB`
`AB=BX` (sides opposite of equal angles)
Now,
`(AB)/(AD)=(BX)/(BC)`
`=>(AB)/(AD)=(BX)/(2BX)` (`X` is mid-point of `BC`)
`=>(AB)/(AD)=1/2`
`therefore AD=2AB`
Hence proved.