Nasty Triangles 5

$\text{ Above sketch is drawn shows the triangle in coordinate plane.}\\ \text{By Law of Cos, } \angle A =Cos^{-1}\dfrac{21^2+15^2 - 24^2}{2*21*15}=81.7868^o.\\ \angle B =Cos^{-1}\dfrac{21^2+24^2 - 15^2}{2*21*24}=38.2132^o.\\ \therefore~Y_{BC}=Tan(90 - B)*X=1.27*X,~~~~~Y_{AC}= Tan(A - 90)*X= - 1.41*X + 21.\\ Y_{BE}=Tan(90 - \dfrac B 2)*X=2.88*X,~~~~~Y_{AD}= Tan(\dfrac A 2 - 90)*X= - 1.15*X + 21.\\ \therefore~Solving ~for~D_X~and~D_Y=1.27*D_X, ~~D(8.66,11).\\ Solving ~for~E_X~and~E_Y=2.88*E_X, ~~E(6.98,20).\\ \therefore~DE^2=(20-11)^2 + (6.98 - 8.66)^2=\Huge~~\color{#D61F06}{84} \\\text{More accurate values were used on calculator to obtain integer 84.}$

By law of cosines: $AB^2=AC^2+BC^2-2AC\times BC \cos C$ $21^2=15^2+24^2-2\times15\times24 \cos C$ $441=225+576-2\times15\times24\cos C$ $2\times15\times24\cos C=801-441=360$ $\cos C=\frac{1}{2}$ $\therefore \angle C=60^\circ$ Since $AD$ is the angle bisector of $\angle BAC$ , we have $\frac{AB}{AC}=\frac{BD}{DC}$ $\frac{BD}{DC}=\frac{7}{5}$ $BD+DC=BC=24$ $\therefore BD=14,~DC=10$ Since $BE$ is the angle bisector of $\angle ABC$ , we have $\frac{AB}{BC}=\frac{AE}{EC}$ $\frac{AE}{EC}=\frac{7}{8}$ $AE+EC=AC=15$ $\therefore AE=7,~EC=8$ Now, by law of cosines, we could find the length of $DE^2$ : $\begin{aligned} DE^2&=EC^2+DC^2-2EC\times DC\cos C \\&=8^2+10^2-8\times10\\&=164-80\\&=84 \end{aligned}$