$ABCD$ is a square. $\angle EAF = 45^o$, $EG$ is perpendicular to $AC$, $FH$ is perpendicular to $AC$, if $AB=5$ and $GH=1$, what is $AG$? 
Since the topic for this question is "cyclic quadrilaterals", I've been looking for cyclic quadrilaterals in this diagram, but haven't found any so far. What I have found is that $\triangle GEC$ and $\triangle FHC$ are 45-45-90, but that isn't much help to me. 
I tried letting $\angle AEG=x$ and $\angle GAF=y$ ($x+y=45$) to angle chase and try to find cyclic quads, but I haven't had much luck. 
 A: From similar triangles AGE and ADF, as well as ABE and AHF,
$$\frac{AG}{AD}=\frac{AE}{AF} = \frac{AB}{AH}\implies 
\frac{AG}5= \frac5{AG+1}$$
which leads to $$AG^2+AG -25=0$$
Solve to obtain
$$AG = \frac12(-1+\sqrt{101})$$
A: Let $\measuredangle CAF=\alpha$.
Thus, $$\measuredangle EAC=\measuredangle DAF=45^{\circ}-\alpha,$$
$$\measuredangle BAE=\alpha,$$
$$AG=AE\cos\alpha=\frac{5\cos\alpha}{\cos(45^{\circ}-\alpha)}.$$
Similarly, $$AH=\frac{5\cos(45^{\circ}-\alpha)}{\cos\alpha}.$$
Now, let $AG=x$.
Thus, $$\frac{25}{x}-x=1.$$
Can you end it now?
A: This is a bit longer than the other answers but the point is to introduce as many cyclic quadrilaterals as possible.
Since $\angle D = \angle AHF = 90^0$, ADFH is cyclic. This means $\angle 1 = \angle 2$.

$\triangle dark green \cong \triangle light green$ implies $\angle 2 = \angle 3$.
As pointed out other answers, $\angle 1 = \angle 4$. Then, $\angle 3 = \angle 4$. [This further implies  EHGZ is cyclic.]
All the blue marked angles are equal to $45^0$. This means ABEG is cyclic. Therefore, $\angle \theta = \angle 4 (= \angle 3)$.
Now, construct the red dotted circle passing through B, G, H. With the fact that $\angle \theta = \angle 3$, we conclude that AB is tangent to the circle BGH.
Let AG = x. By power of a point, $5^2 = x.(x + 1)$, which is the same result as others.   
