IN PARALLELOGRAM ABCD, BISECTOR OF ANGLE A BISECTS BC,WILL BISECTOR OF ANGLE B BISECT AD,EXPLAIN HOW.

Answer 1

I encourage you to sketch parallelogram ABCD and mark it up accordingly as you read through this.

Let point E be the midpoint of BC, and hence also the point where angle A's bisector meets BC.
Let point F be the point where angle B's bisector meets AD.

Draw a line from E to F.
We now have quadrilateral ABEF, with diagonals AE and BF. Because these diagonals bisect angles A and B, respectively (given), then ABEF is a rhombus; i.e., AB = BE = EF = FA.

Now, BE = EC (given), so AB = EC.
We also know that AB = DC ( original parallelogram).
It follows that parallelogram ECDF is also a rhombus, of the same dimensions as rhombus ABEF. Thus F is the midpoint of AD or, in other words, angle B's bisector bisects AD.