Locating the point on the triangle

Question

The distance between $A$ and $B$ is $4$ miles, $A$ and $D$ is $10$ miles, $B$ and $D$ is $7$ miles, and we need to locate the position of the point $C$. A man starts to walk from $A$ to $B$ at a speed of $4$ miles/hour, while another man starts walking from $A$ towards $D$ at $2$ miles/hour. After visiting the point $B$, the first man wants to meet with the second man at a point between $A$ and $D$ such that they both want to meet as soon as possible. Suppose they meet at point $C$, how to locate the position of the point $C$.

What I have tried: Let's say they meet after time $t$, I can use something like this $$\frac{AB + BC}{\text{speed of person $1$}} = \frac{CA}{\text{speed of person 2}}.$$ Then find the equation of line $AD$ assuming I have the coordinates, too. Solve both the equations simultaneously for $C$. But I need a better method.

Here is a picture:

Answer 1

Set: $x=BC$, $y=AC$, $\alpha=\angle BAD$. The two men meet if $x=2y-4$.

From the cosine rule applied to $ABD$ we have $\cos\alpha=67/80$.

Apply now the cosine rule to $ABC$: $$ x^2=4^2+y^2-8y\cos\alpha. $$ Substitute here $x$ and $\cos\alpha$ as given above and solve for $y$.

Answer 2

I don't think there is a shortcut to this:

from this picture: $X=2tsin\alpha=4sin\alpha$

$X^2+(t+t'-2tcos\alpha)^2=(2t')^2$

$16sin^2\alpha+(2+t'-4cos\alpha)^2=4t'^2$

One equation with one unknown, once $\alpha$ known, find $t'$ then $t+t'$.