In triangle $ABC$, the side lengths are $AB = 6,$ $BC = 8,$ and $CA = 8.$ If $AD$ is an angle bisector such that $D$ is on $BC,$ then find the value of $AD^2.$
+177 There are probably many approaches that would lead to the same answer, but here is the approach I decided to use. This problem mentions the existence of an angle bisector, so, naturally. I gravitated towards using the Angle Bisector Theorem to solve this particular problem. We know that \(BC = 8\), so \(BD = BD\), and \(CD = 8 - BD\). In relation to this problem, the Angle Bisector Theorem states \(\frac{AB}{AC} = \frac{BD}{8 - BD}\). Now, I solve for BD.
\(48 - 6BD = 8BD \\ 48 = 14BD \\ BD = \frac{24}{7}\)
Now, we know that \(BD = \frac{24}{7}\). We can apply the Law of Cosines two times to obtain the length of the angle bisector.
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos m\angle B \\ 64 = 36 + 64 - 96\cos m\angle B \\ -96\cos m \angle B = -16 \\ \cos m \angle B = \frac{3}{8}\)
We need not go further than this. We can use this information in the second application of the Law of Cosines.
\(AD^2 = AB^2 + BD^2 - 2(AB)(BD)\cos m \angle B \\ AD^2 = 36 + \frac{576}{49} - 2 * 6 * \frac{24}{7} * \frac{3}{8} \)
At this point, the number were becoming too unwieldy, so I enlisted some help from a calculator.
\(AD^2 = \frac{2340}{49} - \frac{108}{7} = \frac{1584}{49} \approx 32.3265\)
