A triangle has corners at points A, B, and C. Side AB has a length of #35 #. 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 #35  #.  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?

Allison Allen · Accepted Answer

#BC=28.4# Using angle bisector theorem (in the figure drawn by Rui D.), we have #a/b=m/n# or #36/35=m/14# Hence #m=36/35xx14=72/5=14.4# Hence #BC=14.4+14=28.4#

Mason Adams · Answer

The right answer for this problem is given by @shwetank-m (Shwetank Mauria). In explanation I give an answer for the case in which the segment between point A and the intersection with side BC is 14.

Refer to the figure below

Applying the Law of Sines
to #triangle_(ACD)#

#a/sin beta=m/sin alpha#

to #triangle_(ABD)#

#b/sin(180^@-beta)=n/sin alpha# => #b/sin beta=n/sin alpha#

So
#a/b=m/n# => #m=(a/b)n#

NOTE 1
WHAT IS WRITTEN ABOVE STANDS JUST AS PREAMBLE TO THE ANSWER BY @shwetank-m, WHEN -> N=14

NOTE 2
WHAT IS WRITTEN BELOW IS THE ANSWER WHEN -> D=14

Applying the Law of Cosines to both the aforementioned triangles

#m^2=a^2+d^2-2ad*cos alpha# => #m^2-a^2-d^2=-2ad*cos alpha#
#n^2=b^2+d^2-2bd*cos alpha# => #n^2-b^2-d^2=-2bd*cos alpha#

Dividing the two last expressions and substituting #m# we get
#(a^2/b^2n^2-a^2-d^2)/(n^2-b^2-d^2)=a/b#
#a^2/bn^2-b(a^2+d^2)=an^2-a(b^2+d^2)#
#(a^2/b-a)n^2=-a(b^2+d^2)+b(a^2+d^2)#
#a/b(b-a)n^2=a(b^2+d^2)-b(a^2+d^2)#
#n=sqrt(b/a*(a(b^2+d^2)-b(a^2+d^2))/(b-a))#

For #a=35#, #b=36# and #d=14#

#n=sqrt(36/35*(35(36^2+14^2)-36(35^2+14^2))/(36-35))#
#n=sqrt(36/35*(35*1492-36*1421)/1)=sqrt(36/35*1064)=12sqrt(38/5)~=33.08#
=>#m=35/36*12sqrt(38/5)=35/3sqrt(38/5)~=32.16#

So
#BC=c=m+n=35/3sqrt(38/5)+12sqrt(38/5)=71/3sqrt(38/5)~=65.24#

Maverick Smith · Answer

undefined Using the angle bisector theorem, we can determine the length of side BC. According to the theorem, the ratio of the length of side AB to side AC is equal to the ratio of the length of the segment BC to the length of the segment CB'. Here, CB' represents the segment on side BC created by the angle bisector of angle A.

So, we have:

$$\frac{AB}{AC} = \frac{BC}{CB'}$$

Given that AB = 35 and AC = 36, and since BC + CB' = 36 (due to the angle bisector splitting side AC), we can set up an equation:

$$\frac{35}{36} = \frac{BC}{36 - BC}$$

Now, we can solve for BC:

$$35(36 - BC) = 36BC$$
$$1260 - 35BC = 36BC$$
$$1260 = 71BC$$
$$BC = \frac{1260}{71}$$
$$BC ≈ 17.75$$

So, the length of side BC is approximately 17.75 units.