A triangle has corners at points A, B, and C. Side AB has a length of #21 #. The distance between the intersection of point A's angle bisector with side BC and point B is #5 #. If side AC has a length of #18 #, what is the length of side BC?

Let D be the point on BC where the angle bisector, intersects with BC

Then BC = BD+ CD

We know that BD = 5 and have to find CD

Substituting known values into this equation.

Using the Angle Bisector Theorem, we can determine the length of side BC. The theorem states that in a triangle, if an angle bisector intersects the opposite side, it divides that side into segments proportional to the lengths of the adjacent sides.

Let ( D ) be the point where the angle bisector of angle ( A ) intersects side ( BC ). According to the theorem:

[ \frac{BD}{CD} = \frac{AB}{AC} ]

Given ( AB = 21 ) and ( AC = 18 ), we can substitute these values:

[ \frac{BD}{CD} = \frac{21}{18} ]

Now, we can solve for ( BD ) and ( CD ):

[ BD = \frac{21}{18} \cdot CD ]

[ CD = \frac{18}{21} \cdot BD ]

We're also given that the distance between ( D ) and ( B ) is ( 5 ). So, we can write:

[ BD - CD = 5 ]

Substitute the expressions for ( BD ) and ( CD ):

[ \frac{21}{18} \cdot CD - CD = 5 ]

[ \left(\frac{21}{18} - 1\right) \cdot CD = 5 ]

[ \frac{3}{18} \cdot CD = 5 ]

[ \frac{1}{6} \cdot CD = 5 ]

[ CD = 5 \times 6 ]

[ CD = 30 ]

Since ( CD ) represents the length of side ( BC ), ( BC = CD = 30 ).