Example 4 Solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{y-q r}+\sqrt{z-q r}=\sqrt{1-p}, \\
\sqrt{z-p}+\sqrt{x-p p}=\sqrt{1-q}, \\
\sqrt{x-p q}+\sqrt{y-p q}=\sqrt{1-r} .
\end{array}\right.
$$

where $p, q, r$ are positive numbers, and satisfy
$$
p+q+r+2 \sqrt{p q r}=1 .
$$
Solution 1: Treating equation (4) as a quadratic equation in $\sqrt{p}$, we get $\sqrt{p}=\sqrt{(1-q)(1-r)}-\sqrt{q r}$.
Then $1-p$
$$
\begin{array}{c}
=q+r-2 q r+2 \sqrt{q r(1-q)(1-r)}, \\
\sqrt{1-p}=\sqrt{q(1-r)}+\sqrt{r(1-q)} .
\end{array}
$$

Therefore, $(x, y, z)=(p, q, r)$
satisfies equation (1).
Similarly, equation (5) also satisfies equations (2) and (3).
Therefore, equation (5) is one solution to the original system of equations.
If the system of equations has a solution $\left(x^{\prime}, y^{\prime}, z^{\prime}\right), x^{\prime}>p$, then by equation (3) we know $y^{\prime}r$, and by equation (2) we know $x^{\prime}<p$, leading to a contradiction. Similarly, $x^{\prime}<p$ also leads to a contradiction. Therefore, $x^{\prime}=p$. Similarly, $y^{\prime}=q, z^{\prime}=r$. Hence, equation (5) is the unique solution to the original system of equations.

Solution 2: Clearly, $0<p, q, r<1$. Therefore, $0<\sqrt{p}$, $\sqrt{q}, \sqrt{r}<1$. Let three acute angles $A, B, C$ be such that $\sqrt{p}=\cos A$, $\sqrt{q}=\cos B, \sqrt{r}=\cos C$, then equation (4) becomes
$$
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cdot \cos B \cdot \cos C=1 \text {. }
$$

It is easy to prove that equation (6) holds if and only if $A+B+C=\pi$ (proof left to the reader).

As shown in Figure 2, let the circumradius of the acute triangle $\triangle A B C$ be $\frac{1}{2}$, and its pedal triangle be $\triangle D E F$, with the orthocenter $H$, then it is easy to get
$\triangle A E F \backsim$ $\triangle A B C$ (since $B, C, E, F$ are concyclic) $\Rightarrow \frac{E F}{B C}=\frac{A F}{A C}=$ $\cos A$;
$A, F, H, E$ are concyclic $\Rightarrow A H=\frac{E F}{\sin A}$ (sine rule);
$$
\begin{array}{l}
\frac{B C}{\sin A}=1 \Rightarrow \sin A=B C \text { (sine rule). } \\
\therefore A H=\frac{B C \cdot \cos A}{\sin A}=\cos A=\sqrt{P} .
\end{array}
$$

Similarly, $B H=\cos B=\sqrt{q}, C H=\cos C=\sqrt{r}$. Also,
$$
\begin{array}{c}
S_{\triangle B H C}=\frac{1}{2} B C \cdot D H=\frac{1}{2} E C \cdot B H \\
\Rightarrow D H=\cos B \cdot \cos C=\sqrt{q r} .
\end{array}
$$

Similarly, $E H=\cos C \cdot \cos A=\sqrt{P}$,
$$
\begin{array}{l}
F H=\cos A \cdot \cos B=\sqrt{p q} . \\
\text { By } \sqrt{B H^{2}-D H^{2}}+\sqrt{C H^{2}-D H^{2}} \\
=B D+C D=B C \\
=\sin A=\sqrt{1-\cos ^{2} A},
\end{array}
$$

we get $\sqrt{q-q r}+\sqrt{r-q r}=\sqrt{1-p}$.
Similarly, $\sqrt{r-q p}+\sqrt{p-q p}=\sqrt{1-q}$,
$$
\begin{array}{c}
\sqrt{p-p q}+\sqrt{q-p q}=\sqrt{1-r} . \\
\text { Hence }(x, y, z)=\left(A H^{2}, B H^{2}, C H^{2}\right)=(p, q,
\end{array}
$$
$r$ ) is one solution to the original system of equations. The rest is the same as Solution 1.
Figures 1 and 2 help us discover the background and origin of the systems of equations in Examples 3 and 4, especially Figure 2. In fact,
$\triangle A E F \backsim \triangle D B F \backsim \triangle D E C \backsim \triangle A B C$,

the similarity ratios are
$\cos A: \cos B: \cos C: 1$.

Then $S_{\triangle A E F}=\cos ^{2} A, S_{\triangle D B F}=\cos ^{2} B$,
$$
S_{\triangle D E C}=\cos ^{2} C \text {. }
$$

And $S_{\triangle D E F}=2 \cos A \cdot \cos B \cdot \cos C$ (calculation omitted).
The geometric interpretation of equation (6) is
$$
\begin{array}{l}
S_{\triangle A E F}+S_{\triangle D B F}+S_{\triangle D E C}+S_{\triangle D E F} \\
=S_{\triangle A B C} .
\end{array}
$$

This is the source of equation (4). Equations (1), (2), and (3) clearly originate from
$$
\begin{array}{l}
B D+C D=B C, C E+A E=C A, \\
A F+B F=A B .
\end{array}
$$