<s>[INST] Plane geometry tough question $\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.
 [/INST] 
Add up the the sides of $\triangle ADB$ and $\triangle ADC$ and you get their perimeters and then equate them:
$x+\frac{100}{3}-y+\frac{80}{3}=x+y+20$
Solve for $y$ to get $y=20$
and then for $x$ using the cosine rule we get
$x=\sqrt{20^2+20^2-2(20)(20)\cos\left(\tan^{-1}\frac{80}{20\times3}\right)}$
$x=8\sqrt{5}$
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