Vtyjkyuk

I think this is a bit odd but I am juggling since hours with $sin$, $cos$, $tan$ and other stuff to proof a formula, but I can't do it. Slowly I am thinking that this formula is wrong. Maybe there is some expert who could tell me if I am right. I have the following problem:

In the end I want to reach the form:

$$
L_BC = fracL_ACcosalpha - L_AC'sinalpha
$$

starting with the formula for similar triangles:

$$
fracL_ACsintheta = fracL_AC'sin( theta - alpha )
$$

When I combine these two formulas I come to the point that

$$
L_BC = L_AC' fraccosthetasin(theta - alpha)
$$

Now I don't see any way to replace $ theta $ so that I am only dependent on the known variables:
$$
L_AC hspace1cm L_AC' hspace1cm alpha
$$

Also expanding the fractions with sin / cos brings me to an deadend. Am I not seeing an obvious connection in these triangles or is there really something wrong about the formula?
To make the actual question more clear: I want to calculate $L_BC$ using only $alpha$ , $L_AC'$ and $L_AC$! And yes, we have $L_AB=L_AB′$! Thanks!

Let $AB = AB'equiv x$
then
$$ BC = xcostheta
\AC = xsintheta $$

And
$$ AC'=x sin(theta-alpha)
\ implies AC'=xsintheta cosalpha -xcostheta sin alpha
\AC'= AC cosalpha - BCsinalpha
$$

I am going to guess that $AB = AB'$

In which case $AC,AC', BC, BC'$ are proportional to $costheta, cos (theta + alpha),sintheta, sin(theta + alpha)$

And, you are trying to show.

$sin theta = frac costhetacosalpha - cos (theta+alpha)sinalpha$

Which simplifies to
$cos (theta+alpha) = costhetacosalpha - sinthetasinalpha$

Which is one of your basic trig identities.

and $ABC$ and $AB'C'$ are not similar triangles and $frac cos thetasin theta = fraccos (theta + alpha)sin (theta - alpha)$ is incorrect.