How do I understand Pythagorean theorem 1) I understand the formula 
$$\frac{BC}{AB}=\frac{BH}{BC},\ \frac{AC}{AB}=\frac{AH}{AC}$$
But I can't understand the formula is obtained
$$BC^2=AB\times BH \ \text{and}\ AC^2=AB\times AH$$
Why if somebody multiplies $AB * BH = AC^2$?

2) How is the formula obtained?
$$\frac{\sqrt{AB^2-AC^2}}{AC}=\sqrt{1-\left(\frac{AC}{AB}\right)^2}$$
I understand first part, but I can't understand second part 

Each equation of the second pair is obtained from an equation in the first pair by "cross-multiplying,"  or equivalently getting rid of the denominators. 
For example, start from $\frac{BC}{AB}=\frac{BH}{BC}$. Multiply both sides by $(AB)(BC)$. We get $(BC)(BC)=(AB)(BH)$. 
Similarly, from $\frac{AC}{AB}=\frac{AH}{AC}$, by multiplying both sides by $(AB)(AC)$ we get $(AC)(AC)=(AB)(AH)$. 
To finish, take the two equations $BC^2=(AB)(BH)$ and $AC^2=(AB)(AH)$ and add. We get
$$BC^2+AC^2=AB(BH+AH)=(AB)(AB)=AB^2.$$
Remark: As to your question 2, the quoted formula is wrong.  For example, let $AB=5$ and $AC=4$. For these values, the the left-hand side of your expression is equal to $\frac{3}{4}$. The right-hand side is equal to $\frac{3}{5}$.
Added: It turns out that OP wants to prove the correct
$$\frac{\sqrt{AB^2-AC^2}}{AB}=\sqrt{1-\left(\frac{AC}{AB}\right)^2}.\tag{1}$$
Rewrite the left side of (1) as $\sqrt{\frac{AB^2-AC^2}{AB^2}}$. Now divide term by term. We have $\frac{AB^2}{AB^2}=1$  and $\frac{AC^2}{AB^2}=\left(\frac{AC}{AB}\right)^2$, so we obtain the right-hand side of (1)/
The context is that if $\alpha$ is the angle at $A$, then
$\sin\alpha=\frac{BC}{AB}= \frac{\sqrt{AB^2-AC^2}}{AB}$. So on the left-hand side of (1) we have $\sin\alpha$, and on the right-hand side we have $1-\cos^2\alpha$. 
