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 $AD$.
 A: Hint:
You can solve following these steps:

A: 
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
$x=\sqrt{20^2+20^2-2(20)(20)\cos\left(\tan^{-1}\frac{80}{20\times3}\right)}$
$x=8\sqrt{5}$
A: Hint:
$AB+BD+AD=AD+CD+AC$
$AB+BD=CD+AC$
$AB+BD+DC=2CD+AC$
$20+\dfrac{100}{3}=2CD+\dfrac{80}{3}$
Solve for $CD$ .And then for $AD$.
A: Referring to zewzad's great sketch, I used Maazul's algebra to determine that
BD = 20
I then used coordinate vector geometry to obtain the coordinates of point D (defining the origin at A).  This leads to coordinate values:
A = [ 0 , 0 ]
B = [ 0,  20 ]
C = [ 80/3, 0 ]
I used scalar multiplication and vector addition to obtain coordinates for D.
The vector that takes you from B --> C is < 80/3, -20 >.  We know where point D lies fractionally going from B --> C.  It's (20 / 100/3) or 0.6 of the full distance.  Scalar-mult vector B-->C by 0.6 to get the vector B-->D:
B-->D = 0.6 * B-->C
B-->D = 0.6 * < 80/3, -20 >
  = < 16 , -12 >

by vector addition, point D has coordinates:
D = B + B-->D
= [ 0, 20 ] + < 16 , -12 >
= [ 16, 8 ]
Finally, distance AD can be computed as length of point-vector D = ( <16, 8> )
AD   = sqrt (16^^2 + 8^^2 )
AD   = sqrt (320)
AD   = 17.8888    = 8*sqrt(5) 
