An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.

I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
with $0<\beta<\pi$. Hence $\sin \beta=\sqrt{1-5/25}=2\sqrt5/5$. But $\overline{AB}=\overline{BC}\cos \beta$ and the area
$$\mathcal A(\triangle ABC)=\frac12(\overline{BC}\cos \beta)\cdot (\overline{BC}\sin \beta)=\frac12\overline{BC}^2 \tag 1$$
We know that:
$$\overline{AH}+\overline{CH}+\overline{HB}=7, \quad \text{with}\quad  \overline{CH}+\overline{HB}=\overline{BC}$$
Hence
$$\overline{AH}+\overline{BC}=7 \iff \overline{BC}=7-\overline{AH}$$
and $\overline{AH}=\overline{AB}\cos\gamma=\overline{BC}\cos\gamma\cos\beta$.
Putting this last identity to the condition $\overline{BC}=7-\overline{AH}$ I will have:
$$\overline{BC}=7-\overline{BC}\cos\gamma\cos\beta \iff \overline{BC}(1+\cos\gamma\cos\beta)=7$$
After,
$$\overline{BC}=\frac{7}{1+\cos\gamma\cos\beta}$$
with $\cos \gamma=\cos(\pi/2-\cos\beta)=\sin\beta=2\sqrt5/5$.
Definitively
$$\overline{BC}^2=\frac{49}{\left(1+\frac 25\right)^2}=\frac{49}{\frac{49}{25}}$$
and
$$\mathcal{A}(\triangle ABC)=\frac15\overline{BC}^2=5 \tag 2$$
I'm very tired and often don't find immediate solutions to problems. I had thought of using Euclid's second theorem by placing
$\overline{AH}=x$ and $\overline{BC}=y$, and $x+y=7$. After I will have $x^2=\overline{BH}\cdot \overline{CH}$. But I have left this path.

Can this possible alternative be used or is there another more immediate way?


$$\cos \beta = \frac{|AB|}{|BC|} = \frac{\sqrt 5}{5},\quad  \sin \beta = \frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}$$
Given $|AB|: |BC| = \sqrt5:5$, we assume $$|AB| = \sqrt5k, |BC| = 5k $$ where $k$ is a constant. Now using $$\frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}, \quad |AH| = 2k$$
$$|AH| + |BH| + |HC| = |AH| + |BC| = 7k = 7\implies k = 1.$$
So, $$|AH| = 2, |BC| = 5 \implies \mathcal{A}(\triangle ABC) = \frac{1}{2} \cdot |AH| \cdot |BC| = 5.$$
