A triangle has corners at points A, B, and C. Side AB has a length of #18 #. 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 #21 #, what is the length of side BC?

Question

A triangle has corners at points A, B, and C.  Side AB has a length of #18  #.  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 #21 #, what is the length of side BC?

Violet Aldrich · Accepted Answer

#BC=6.5#

Please refer to figure below.

Here let #a=BC#,
#b=AC=21# and
#c=AB=18#.

Further, bisector of angle #A# cuts #AB# at #D# and #BD=3#

In such a triangle according to angle bisector theorem, bisector of angle #A#, divides #BC#

in the ratio of the two sides containing the angle.

In other words, #(AB)/(AC)=(BD)/(DC)# and hence here we have

#18/21=3/(DC)# or #18xxDC=21xx3# and

#DC=(21xx3)/18=(7cancel(21)xx3)/(6cancel(18))=7xx3/6=7/2=3.5#

Hence #BC=3+3.5=6.5#

Sadie Adams · Answer

undefined Using the Angle Bisector Theorem, the length of side BC can be determined. The theorem states that in a triangle, if a line bisects one of the angles, it divides the opposite side into segments proportional to the other two sides.

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

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

Given that AB = 18, AC = 21, and BD = 3, we can plug these values into the equation and solve for DC:

$$ \frac{3}{DC} = \frac{18}{21} $$

$$ \frac{3}{DC} = \frac{6}{7} $$

$$ DC = \frac{7}{2} $$

Now, we can find the length of BC by adding BD and DC:

$$ BC = BD + DC $$
$$ BC = 3 + \frac{7}{2} $$
$$ BC = \frac{13}{2} $$

Thus, the length of side BC is $ \frac{13}{2} $ units.