Irish Mathematical Olympiad, 2014

Because $A, B, C, F$ are on a circle, we have $\angle BFC = \angle BAC = 45^\circ$. Since $\angle AFC = 90^\circ$ this means that $FB$ is the bisector of the angle $\angle AFE$. Because the angle bisector cuts the opposite side in a triangle at a ratio equal to the ratio of the adjacent sides (which follows easily from the Sine Theorem or by calculating areas in two different ways), we obtain
$$\frac{|AH|}{|HE|} = \frac{|AF|}{|FE|}$$

Because $\angle AFE = \angle AFC = 90^\circ = \angle EDC$ and $\angle FEA = \angle DEC$, the triangles AFE and CDE are similar, so that $\frac{|AF|}{|FE|} = \frac{|CD|}{|DE|} = 2$.
We obtain now that $|AH| = 2|HE|$. As $|ED| = |AE| = |AH| + |HE| = 3|HE|$, we finally get $|HD| = |HE| + |ED| = 4|HE| = 2|AH|$, as required.

Because $A, B, C, F$ are on a circle, we have $\angle HFE = \angle BFC = \angle BAC$, $\angle AFH = \angle AFB = \angle ADB$ and $\angle EFD = \angle CFD = \angle CAD$, and all these angles are equal to $45^\circ$. With the abbreviations $s = \sin(45^\circ) = \sin(135^\circ)$, $t = \sin(\angle FHA) = \sin(\angle FHE)$ and $u = \sin(\angle FDE)$, the Sine Theorem on the triangles AFH, HEF, EFD and ADF gives
$$\frac{|AF|}{t} = \frac{|AH|}{s}, \quad \frac{|FE|}{t} = \frac{|HE|}{s}, \quad \frac{|FE|}{u} = \frac{|ED|}{s}, \quad \frac{|AF|}{u} = \frac{|AD|}{s}.$$
From these we get
$$\frac{|AH|}{s} \cdot \frac{s}{|HE|} = \frac{|AF|}{t} \cdot \frac{t}{|FE|} = \frac{|AF|}{u} \cdot \frac{u}{|FE|} = \frac{|AD|}{s} \cdot \frac{s}{|ED|} = 2,$$
hence $|AH| = 2|HE|$ and we conclude as in Solution 1.

Using $\tan(45^\circ) = 1$ and the addition theorem for tan, we obtain
$$\frac{\tan(\angle ABF) + \tan(\angle FCD)}{1 - \tan(\angle ABF) \tan(\angle FCD)} = \tan(\angle ABF + \angle FCD) = 1.$$
As $\tan(\angle FCD) = \frac{|ED|}{|CD|} = \frac{1}{2}$, we get $1 - \frac{1}{2}\tan(\angle ABF) = \frac{1}{2} + \tan(\angle ABF)$ and so
$$\frac{1}{3} = \tan(\angle ABF) = \frac{|AH|}{|AB|}$$
This implies $3|AH| = |AB| = |AD| = |AH| + |HD|$ and so $2|AH| = |HD|$.

Considering the power of the points $E$ and $H$ we obtain
$$\begin{aligned} |FE| \cdot |EC| &= |AE| \cdot |ED| \\ |FH| \cdot |HB| &= |AH| \cdot |HD|. \end{aligned}$$
Because $AD$ is parallel to $BC$ we also obtain $\frac{|FE|}{|EC|} = \frac{|FH|}{|HB|}$. This gives
$$\frac{|AE| \cdot |ED|}{|EC|^2} = \frac{|FE|}{|EC|} = \frac{|FH|}{|HB|} = \frac{|AH| \cdot |HD|}{|HB|^2}.$$
By the Theorem of Pythagoras we have
$$|EC|^2 = |CD|^2 + |ED|^2 \quad \text{and} \quad |HB|^2 = |AB|^2 + |AH|^2.$$
Writing $a = |AB| = |CD| = |AD|$ as well as $|AH| = x a$ with $0 < x < \frac{1}{2}$ and using $|AE| = |ED| = a/2$, we obtain
$$\frac{a^2/4}{5a^2/4} = \frac{xa(a - xa)}{a^2 + x^2a^2}$$
which is equivalent to the quadratic equation $6x^2 - 5x + 1 = 0$. Its two roots are $x = 1/2$ and $x = 1/3$. As $x < 1/2$ we must have $x = 1/3$ and this means $3|AH| = |AD|$. We finally obtain $|HD| = |AD| - |AH| = 2|AH|$.