<s>[INST] Geometry problem related to right angled triangles. 
In the given figure $AC = 12 cm, AE = 6 cm$ and $CD = 8 cm$. CD is perpendicular to $AD$ and $BE$ is perpendicular to $AC$. How can we find the value of $BE$?
 [/INST] This is a method without using similar triangles:
First of all, we can find $\text{AD}$ by using the Pythagorean Theorem:
$$\text{AD}=\sqrt{\text{AC}^2-\text{CD}^2}\to\text{AD}=\sqrt{12^2-8^2}=4\sqrt{5}\space\text{cm}$$
So, for $\text{DE}$:
$$\text{DE}=\text{AD}-\text{AE}\to\text{DE}=\left(4\sqrt{5}-6\right)\space\text{cm}$$
Now, we can find $\text{CE}$ using again the Pythagorean Theorem:
$$\text{CE}=\sqrt{\text{CD}^2+\text{DE}^2}\to\text{CE}=\sqrt{8^2+\left(4\sqrt{5}-6\right)^2}=2\sqrt{45-12\sqrt{5}}\space\text{cm}$$
So, we also can say that:
$$
\begin{cases}
\text{BE}^2+\left(12-\text{AB}\right)^2=180-48\sqrt{5}\\
\text{BE}^2+\text{AB}^2=36
\end{cases}\Longleftrightarrow
\begin{cases}
\text{BE}^2=180-48\sqrt{5}-\left(12-\text{AB}\right)^2\\
\text{BE}^2+\text{AB}^2=36
\end{cases}
$$
Now, we can conclude:
$$\left(180-48\sqrt{5}-\left(12-\text{AB}\right)^2\right)+\text{AB}^2=36\Longleftrightarrow\text{AB}=2\sqrt{5}\space\text{cm}$$
So, for $\text{BE}$ we find:
$$\text{BE}^2=180-48\sqrt{5}-\left(12-2\sqrt{5}\right)^2=16\to\text{BE}=4\space\text{cm}$$
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