A triangle has corners at points A, B, and C. Side AB has a length of #15 #. 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 #16 #, what is the length of side BC?

Question

A triangle has corners at points A, B, and C.  Side AB has a length of #15  #.  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 #16  #, what is the length of side BC?

Evelyn Arboleda · Accepted Answer

Length of side BC = 10.3333

Let the point where the angle bisector intersects with side BC be D #"using the "color(blue)"angle bisector theorem"# #(AB)/(AC)=(BD)/(DC)# #15 / 16 = 5 / (DC)# #DC = (5*16) / 15 = 5.3333# #BC = BD+DC= 5+5.3333 =10.3333#

Wyatt Anderson · Answer

undefined To find the length of side BC in the triangle with corners at points A, B, and C, where side AB has a length of 15, the distance between the intersection of point A's angle bisector with side BC and point B is 5, and side AC has a length of 16:

Let's denote the length of side BC as $ x $.

Using the Angle Bisector Theorem, we know that the angle bisector of angle A divides side BC into segments proportional to the other two sides of the triangle. Therefore, we can set up the following proportion:

$$
\frac{AB}{AC} = \frac{BD}{CD}
$$

Plugging in the given values, we have:

$$
\frac{15}{16} = \frac{x}{x + 5}
$$

Cross-multiplying yields:

$$
15(x + 5) = 16x
$$

Expand and solve for $ x $:

$$
15x + 75 = 16x
$$
$$
75 = 16x - 15x
$$
$$
75 = x
$$

So, the length of side BC is 75 units.