Example $1$, $, $ Given line segments $b, c$. Find the line segment $a$ such that
$$
\frac{1}{a}=\frac{1}{b}+\frac{1}{c} .
$$
Analysis: Consider $\mathrm{a}$ as $\mathrm{f}$, $\mathrm{b}$ as $\mathrm{u}$, and $\mathrm{c}$ as $\mathrm{v}$. Using the above physical construction method, we can construct $\mathrm{a}$.
Construction:
(1) Draw a line 1, take a point 0 on 1, and on both sides of 0 on 1, mark $O D=b$ and $O E=c$;
(2) Draw perpendiculars DM, OT, and EN from D, O, and E to 1, and take an appropriate length for DM;
(3) Extend MO to intersect EN at N;
(4) Draw MT $\downarrow \mathrm{TO}$ at T;
(5) Connect $\mathbf{T N}$ to intersect 1 at $\mathrm{F}$.
Then OF is the desired length (Figure 2).
This construction method is more complex than the usual algebraic method, which is to construct $\mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}$. However, it is more intuitive and does not require algebraic manipulation. Is there a simpler construction method? Yes, but we need to prove a proposition first: If points $\mathrm{C}$ and $\mathrm{B}$ are collinear, then $\frac{1}{\mathrm{OC}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}$.
Proof: Extend $\mathrm{BO}$ to $\mathrm{D}$ such that $\mathrm{OD}=\mathrm{OA}$, and connect $\mathrm{DA}$.
Since $\angle \mathrm{AOD}=180^{\circ}-\angle \mathrm{AOB}=60^{\circ}$ and $\mathrm{OD}=\mathrm{OA}$, $\triangle O A D$ is an equilateral triangle.
- Therefore, $\angle \mathrm{ADO}=60^{\circ}=\frac{1}{2} \angle \mathrm{AOB}=\angle \mathrm{CUB}$, so OC is parallel to AD.
Thus, $\triangle B C O \sim \triangle B B$,
$$
\begin{array}{l}
\frac{O B}{B D}=\frac{O C}{A D}, \quad \frac{O B}{O B+O A}=\frac{O C}{O A}, \\
O C=\frac{O A \cdot O B}{O A+O B} . \\
\end{array}
$$

Therefore, $\frac{1}{O C}=\frac{1}{O A}+\frac{1}{O B}$.
According to this proposition, the construction for Example 1 is easily solved.