### Instruction: 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 the value of $AD^2.$ Express your answer as a common fraction.### Response: I know that an angle bisector divides the opposite side into two segments that are proportional to the adjacent sides.
So, in this case, $BD/DC = AB/AC = 5/4.$
That means I can write $BD = 5x$ and $DC = 4x$ for some positive number $x.$
I also know that $BD + DC = BC = 5,$ so I can solve for $x$ by adding the two equations and getting $9x = 5.$
That implies $x = 5/9.$
So, $BD = 5x = 25/9$ and $DC = 4x = 20/9.$
Now, I want to find $AD^2.$
I can use the angle bisector theorem again, but in a different way.
I know that $AD^2 = AB \cdot AC - BD \cdot DC.$
This is a useful formula that relates the length of the angle bisector to the lengths of the sides of the triangle.
Plugging in the values, I get $AD^2 = 5 \cdot 4 - (25/9) \cdot (20/9).$
Simplifying, I get $AD^2 = 20 - 500/81.$
I need to find a common denominator to subtract the fractions.
The least common multiple of 1 and 81 is 81, so I multiply the first term by 81/81.
I get $AD^2 = (20 \cdot 81)/81 - 500/81.$
Combining the fractions, I get $AD^2 = (1620 - 500)/81.$