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?

Question

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?

Sadie Allen · Accepted Answer

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 #

Caroline Aucoin · Answer

undefined Using the angle bisector theorem, we can find the length of side BC. Let D be the point where the angle bisector of angle A intersects side BC.

According to the angle bisector theorem:
$$ \frac{BD}{DC} = \frac{AB}{AC} $$

Given: AB = 45, AC = 42, and BD = 6.
$$ \frac{6}{DC} = \frac{45}{42} $$
$$ \frac{6}{DC} = \frac{15}{14} $$
$$ 6 \times 14 = 15 \times DC $$
$$ 84 = 15 \times DC $$
$$ DC = \frac{84}{15} $$
$$ DC = 5.6 $$

Now, we can find the length of BC using the fact that BC = BD + DC.
$$ BC = 6 + 5.6 $$
$$ BC = 11.6 $$

Therefore, the length of side BC is 11.6.