A triangle has corners at points A, B, and C. Side AB has a length of #12 #. The distance between the intersection of point A's angle bisector with side BC and point B is #3 #. If side AC has a length of #9 #, what is the length of side BC?

Question

A triangle has corners at points A, B, and C.  Side AB has a length of #12  #.  The distance between the intersection of point A's angle bisector with side BC and point B is #3  #.  If side AC has a length of #9  #, what is the length of side BC?

Savannah Aguilar · Accepted Answer

Length of the side #BC# is #5.25# unit.

Sides of triangle are #AB=12 , AC=9 , BC= ?# Let the angle bisector of #/_A#, AD meets BC at D. #BD=3#. By the Angle Bisector Theorem we know, #(BD)/(DC)=(AB)/(AC) or 3/(DC)=12/9 or DC=(3*9)/12#or #DC=9/4 =2.25 :. BC=BD+DC=3+2.25=5.25# #:. BC=5.25# unit . Length of the side #BC# is #5.25# unit [Ans]

Parker Allen · 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 that AB = 12 and AC = 9:

$$\frac{{BD}}{{DC}} = \frac{{12}}{{9}} = \frac{4}{3}$$

Let's denote the length of BD as x. Then DC = (3/4)x.

We're also given that the distance between D and B is 3.

So, x + (3/4)x = 3

Solving for x:

$$x + \frac{3}{4}x = 3$$
$$\frac{7}{4}x = 3$$
$$x = \frac{3 \times 4}{7}$$
$$x = \frac{12}{7}$$

Now, the length of BC is BD + DC:

$$BC = x + \frac{3}{4}x$$
$$BC = \frac{12}{7} + \frac{3}{4} \times \frac{12}{7}$$
$$BC = \frac{12}{7} + \frac{9}{7}$$
$$BC = \frac{21}{7}$$
$$BC = 3$$

So, the length of side BC is 3.