A triangle has corners at points A, B, and C. Side AB has a length of #42 #. The distance between the intersection of point A's angle bisector with side BC and point B is #14 #. If side AC has a length of #36 #, what is the length of side BC?

Question

A triangle has corners at points A, B, and C.  Side AB has a length of #42  #.  The distance between the intersection of point A's angle bisector with side BC and point B is #14  #.  If side AC has a length of #36 #, what is the length of side BC?

Evelyn Arboleda · Accepted Answer

#abs(BC)=26#

Labeling the intersection of A's bisector with side BC as D:
#color(white)("XXX")abs(DC)/36=14/42#

#color(white)("XXX")abs(DC)=(cancel(14)^2xxcancel(36)^6)/(cancel(42)_cancel(6))=12#

#abs(BC)=abs(BD)+abs(DC)#
#color(white)("XXX")=14+12=26#

Maverick Smith · Answer

undefined This problem can be solved using the Angle Bisector Theorem, which states that the angle bisector of an angle in a triangle divides the opposite side into two segments that are proportional to the adjacent sides.

Let $D$ be the point on $BC$ where the angle bisector from $A$ intersects $BC$. Given that $AD$ is the angle bisector, $AB = 42$, $AC = 36$, and $BD = 14$, we need to find $BC$.

According to the Angle Bisector Theorem,

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

Given $AB = 42$, $AC = 36$, and $BD = 14$, let $DC = x$. So,

$$\frac{42}{36} = \frac{14}{x}$$

Simplifying $\frac{42}{36}$ gives $\frac{7}{6}$, so,

$$\frac{7}{6} = \frac{14}{x}$$

Solving for $x$,

$$7x = 14 \times 6$$

$$x = 2 \times 6 = 12$$

So, $DC = 12$.

The length of $BC$ is $BD + DC = 14 + 12 = 26$.

Therefore, the length of side $BC$ is $26$.