We have a triangle $\triangle ABC$ such that $AB = BC = 5$ and $AC = 4.$ If $AD$ is an angle bisector such that $D$ is on $BC,$ then find th

Probably more elegant ways to do this.....but....

Here's a diagram :

AD  will bisect BAC.......and this angle bisector  creates the following ratio

BA / AC  =  BD / DC

5/4  =  BD / DC

Let  DC  =  x     then BD  = (5/4)x

And  DC + BD  = 5        so

x + (5/4)x  = 5

(9/4)x  =  5

x  = 20/9  =  DC

Draw  altitude  BG to AC .....this will bisect AC....   and  draw DF perpendicular to AC

Now......BG  =  √[BC^2 - GC^2]  = √[ 5^2 - 2^2]  = √21

And   triangles BGC  and DFC  are similar

And  BG/ BC   is similar to  DF/DC......so

√21 / 5  =  DF/ (20/9)     so

(√21 / 5) * (20/9)  =  DF   = 4√21 / 9  

Likewise

GC /  BC  =  FC /DC

2/5  =  FC / (20/9)

(20/9)(2/5)  = FC  =  40/45  = 8/9

And AF  =  AC  - FC  =   4  -  8/9     =  28/9

But  ADF  forms a right triangle  with   AF^2  + DF^2   =  AD^2

AF^2  =  (28/9)^2  =   28^2 / 81  =  784/81

DF^2  = [ 4√21 / 9 ]^2  =   16*21 / 81  =  336/81

So    AD^2  =    [ 336 + 784 ] 81  =  1120/81

In triangle ABD, angle BAD =33.2 deg., Angle ABD=47.2 deg. Angle ADB =99.6 deg.
Side AB = 5
Using the Law of Sines, we get:BD=2.777 and AD=3.721, so:
AD^2 =3.721^2 =13.846 ???!!
CPhill: Does this make sense???

Similarly, in triangle ACD, angle DAC=33.2 deg, angle DCA=66.4 deg, and angle CDA=80.4 deg.

Side AC=4.  By Law of Sines: CD=2.221 and AD=3.718. Therefore:

AD^2 =3.718^2 =13.823 ~ 1,120/81.