Trigonometry and algebra question Given:


*

*The total length of ad + dc

*The lengths of each ab, bc and ca

*That angle adb = angle cdb
How can I work out lengths ad and cd

If you can't see the image above, click here
 A: Geometric Solution:
Let $h=\dfrac{|\overline{bc}|}2$ and $k=\dfrac{|\overline{bd}|+|\overline{cd}|}2$.
Place $c$ at $(-h,0)$ and $b$ at $(+h,0)$. $a$ would then be at the intersection of the circle of radius $|\overline{ac}|$ centered at $c$ and the circle of radius $|\overline{ab}|$ centered at $b$.
Since $|\overline{bd}|+|\overline{cd}|=2k$ (i.e. a constant), $d$ lies on the ellipse
$$
\frac{x^2}{k^2}+\frac{y^2}{k^2-h^2}=1\tag{1}
$$
Graphically,
$\hspace{3.2cm}$
Because of the reflection property of ellipses, $\angle adc=\angle adb$ when $\overline{ad}$ is normal to the ellipse (then $\angle adc$ and $\angle adb$ are supplementary to the angle of incidence). That is, $d$ is the point on the ellipse given in $(1)$ closest to (or furthest from) $a$.
The slope of the normal to the ellipse at $d=(x,y)$ is
$$
\frac yx\frac{k^2}{k^2-h^2}\tag{2}
$$
Therefore, we need
$$
\frac{y-a_y}{x-a_x}=\frac yx\frac{k^2}{k^2-h^2}\tag{3}
$$
which becomes the right hyperbola
$$
\left(x-\frac{k^2}{h^2}a_x\right)\left(y+\frac{k^2-h^2}{h^2}a_y\right)
=-\frac{k^2}{h^2}\frac{k^2-h^2}{h^2}a_xa_y\tag{4}
$$
Thus, we can find $d$ at the intersection of the ellipse in $(1)$ and the right hyperbola in $(4)$.

For computation of $d$, it is probably easiest to use
$$
u=x-\frac{k^2}{h^2}a_x\quad\text{and}\quad v=y+\frac{k^2-h^2}{h^2}a_y\tag{5}
$$
Then, we have from $(1)$ and $(4)$,
$$
\frac{\left(u+\frac{k^2}{h^2}a_x\right)^2}{k^2}+\frac{\left(v-\frac{k^2-h^2}{h^2}a_y\right)^2}{k^2-h^2}=1\tag{6}
$$
and
$$
uv=-\frac{k^2}{h^2}\frac{k^2-h^2}{h^2}a_xa_y\tag{7}
$$
Using $(7)$ in $(6)$ yields a fourth degree equation to solve.
