Name this triangle ABC, where A is the angle divided into equal parts by bisector AD and point B lies under the bisector AD and point C lies over the bisector AD.
According to the bisector theorem:
[tex] \dfrac{AB}{AC} = \dfrac{BD}{BC} [/tex].
On the picture CD=q and BC=16, then BD=16-q.
The previous ratio becomes:
[tex] \dfrac{28}{36} = \dfrac{16-q}{q}\\ 28q=36(16-q)\\ 28q=576-36q\\ 64q=576\\q=9 [/tex].
Answer: q=9
