What is length of $AD$ in this triangle?

Question

I'm preparing for a test and was not able to solve.

Here's the question:

From the corner $C$ of the triangle $ABC$ draw $CD$, such that $∠ BCD = ∠ BAC$ and $D$ lies on $AB$. Calculate $|AD|$, if $AB = 17$ and $BC = 10$.

That is how the picture may look like:

We didn't have law of sines or cosines yet. But it can probably be found somewhere so I don't yet know how to use it. From just googling I don't see how this could be solved without the angle even though they are the same.

Now, I tried using the fact that angles are the same could the side lengths also be the same

Nope, that surely can't be as one side is used to make the other, else, it would have to have two sides the same(I don't know how that is called in English)

Answer 1

Rather than using law of sines/cosines as suggested in the comments you could use AA similarity between $BCD$ and $BAC$ (they are similar by the statement given and the reflexive property). Call AD x, then BD is $17-x$. From the similar triangles mentioned earlier, $$\frac{10}{17-x}=\frac{17}{10}$$ Which you can solve to get $x = \frac{189}{17}$

Answer 2

Let $AD=x$.

Since $\Delta ACB\sim\Delta CDB$, we obtain: $$\frac{17-x}{10}=\frac{10}{17},$$ which gives $x=\frac{189}{17}.$