For reference:In the picture shown we have an isosceles trapezoide,
The $QR$ side measures $b$ and the $PQ$ side measures $a$. find $LR$
(answer:$\frac{b(a+b)}{a-b}$ )
My progress:
I didn't get much on this question... a colleague indicated if PQRL form a harmonic range. we would have a quick solution. But it would be necessary to demonstrate that these points form a harmonic range ou or if someone sees another solution...
$LR = x\\ \frac{a+b+x}{x}=\frac{a}{b}\\ ax=ba+b.b+bx\\ ax - bx=b(a+b) \\ x(a-b) = b(a+b)\\ \therefore x = \frac{b(a+b)}{(a-b)}$
Edit: I found a way to demonstrate that points L, R, Q and P form a harmonic range:
$\angle DCP = 2\theta\\\angle DCB = 2\alpha\\\therefore \boxed{\alpha + \theta = 90^o}\\O ~is~ excenter~ \triangle ARL\\\angle LOR = \alpha\\O ~is~ excenter~ \triangle QCP\\\angle QOP = \theta\\\\ \angle ROQ = 90^o-\theta = \alpha \\ \triangle LOQ:\\ OR \text{ is angle bissector internal}\\ \text{OP is angle bissector external} $
Showing the pencil $O (LQ; RP)$ is harmonic definitely makes the work easier. Here is an alternate approach that requires solving a quadratic equation in the end.
$EF$ is perpendicular bisector of $AD$ and $BC$. In $EFRP$, $\angle FRP = 180^\circ - \angle EPR = 180^\circ - 2\theta$
As $F$ and $T$ are points of tangency on the circle from external point $R$, $OR$ bisects $\angle FRT$ and $\angle ORT = 90^\circ - \theta$
So, $\triangle OTR \sim \triangle PTO$. If $RT = c$ and $r$ is the radius of the circle, we have
$ \displaystyle \frac{c}{r} = \frac{r}{a+b-c} \implies r^2 = c (a+b-c) \tag1$
$\angle CQR = \angle PCQ + \angle CPQ = 180^\circ - 2 \alpha + 2\theta$
As $OQ$ bisects $\angle CQR$, $\angle OQT = 90^\circ + \theta - \alpha$ and it follows that $\angle ROQ = \alpha$
Using your work, $\angle LOR = \alpha$ or alternatively, $\angle OLT = \frac{180^\circ - (2 \alpha + 2\theta)}{2} = 90^\circ - (\alpha + \theta)$
SO, $\angle LOT = \alpha + \theta \implies \angle LOR = \alpha$. So $OR$ is bisector of $\angle LOQ$.
Therefore, $ \displaystyle \frac{OL^2}{OQ^2} = \frac{LR^2}{RQ^2}$
If $LR = x, ~ \displaystyle \frac{(x+c)^2 + r^2}{r^2 + (b-c)^2} = \frac{x^2}{b^2}$
Or, $b^2 c^2 + 2b^2cx + b^2 r^2 = x^2 ( r^2 + c^2 - 2 bc)$
Plugging in value of $r^2$ from $(1)$ and dividing both sides by $c$,
$b^2 (a+b) + 2b^2 x = (a-b) x^2$
$\implies (a-b) x^2 - 2b^2 x - b^2 (a+b) = 0$
$ ((a-b)x - b(a+b)) (x + b) = 0$
That leads to, $ \displaystyle x = \frac{b (a + b)}{a-b}$