In `DeltaABC`, ray `BD` bisects `/_ABC`. `A-D-C`, side `DE||` side `BC`, `A-E-B`.
Prove that, `(AB)/(BC)=(AE)/(EB)`.
Complete the activity by filling the boxes.

In `DeltaABC`, ray `BD` is the bisector of `/_ABC`
`:.(AB)/(BC)=square`.......`(I)` (By angle bisector theorem)
In `DeltaABC`, seg `DE||` side `BC`
`:.(AE)/(EB)=(AD)/(DC)`........`(II)` `square`
`:.(AB)/(square)=(square)/(EB)`.......[From `(I)` and `(II)`]

Explanation

In `DeltaABC`, ray `BD` is the bisector of `/_ABC`
`:.(AB)/(BC)=(AD)/(DC)`.......`(I)` (By angle bisector theorem)
In `DeltaABC`, seg `DE||` side `BC`
`:.(AE)/(EB)=(AD)/(DC)`........`(II)` By basic proportionality theorem
`:.(AB)/(BC)=(AE)/(EB)`.......[From `(I)` and `(II)`]