For reference (exact copy of the question):
In the $ABCD$ quadrilateral circumscribed to a circle, $AB$ and $CD$ are tangents at $T$ and $Q$ respectively, $TQ$ intercepts $AC$ at $E$. $EC = 10$, $QC = 8$ and $AT = 4$.
Calculate $AE$.
My progress:
It follows the resolution by trigonometry... could anyone solve it by geometry?
$\angle EQC=\alpha$
$OQ$ is ratio, therefore $OQ \perp QC \Rightarrow \theta +\alpha=90^\circ$.
$\triangle OQT$ isosceles, therefore $\angle OQT=\angle OTQ$, therefore $\angle QTB=\alpha$.
Law of sines:
- $\triangle ATE$:
$$ \tag{$I$} \frac{AT}{\sin \beta} =\frac{AE}{{\underbrace{\sin (\pi−\alpha)}}_{\sin\alpha}}\ . $$
- $\triangle ECQ$:
$$ \tag{$II$} \frac{QC}{\sin\beta}=\frac{CE}{\sin\alpha} \ . $$
$(II) \div (I)$ gives: $$\frac{AT}{QC}=\frac{AE}{CE} \qquad\Rightarrow\qquad \frac{4}{8}=\frac{AE}{10}\ , $$ and therefore $AE=5$.
The points $B,D$ are not needed, so let us remove them from the picture. Now the circle plays only the rôle of making the two angles in $T,Q$, marked in the posted picture by the same letter $\alpha$ to have the same measure. So let us isolate the essence, the problem is reduced to the following one:
As shown in the picture, let $T^*$ be on $QE$ such that $\Delta ATT^*$ is isosceles (in $A$). (For this construct $T^*$ as the reflection of $T$ w.r.t. the perpendicular line on $QET$ which is passing through $A$.)
It is now easy to see the similarity: $$ \Delta AT^*E\sim \Delta CQE\ , $$ (corresponding two angles of measure $\alpha,\beta$), so we get: $$ \frac {AT}{AE} = \frac {AT^*}{AE} = \frac {CQ}{CE} \ . $$ $\square$