A triangle has corners at points A, B, and C. Side AB has a length of #45 #. The distance between the intersection of point A's angle bisector with side BC and point B is #6 #. If side AC has a length of #42 #, what is the length of side BC?

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Answer 1 · 2018-04-23T12:59:29

By the angle bisector theorem, I get #BC=58/5#.

We call the corners "vertices" -- it sounds smarter.

Let's call the foot of A's angle bisector D. That's where it meets BC.

We're given #AB=45, AC=42, BD=6#.

By the angle bisector theorem, we have proportionality

#{AC}/{CD} = {AB}/{BD} #

# CD = frac{AC \ BD}{AB} = { (42)(6) }/45 = 28/5 #

#BC = BD + CD = 6 + 28/5 = 58/5 #