76 Two perpendicular lines passing through the origin intersect the parabola $y^{2}=4 p(x+p)(p>0)$ at $A, B$ and $C, D$. When is $|A B|+|C D|$ minimized?
76 Let the lines $AB$ and $CD$ be
$$
\begin{array}{c}
y=m x, \\
y=-\frac{1}{m} x .
\end{array}
$$

From (1) and $y^{2}=4 p(x+p)$, we get
$$
m^{2} x^{2}=4 p(x+p),
$$

which simplifies to
$$
m^{2} x^{2}-4 p x-4 p^{2}=0 \text {. }
$$

Let the coordinates of $A$ and $B$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. By Vieta's formulas, we have
$$
x_{1}+x_{2}=\frac{4 p}{m^{2}}, x_{1} \cdot x_{2}=\frac{-4 p^{2}}{m^{2}} \text {, }
$$

Thus,
$$
\begin{aligned}
A B^{2} & =\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(m x_{1}-m x_{2}\right)^{2} \\
& =\left(m^{2}+1\right)\left(x_{1}-x_{2}\right)^{2}=\left(m^{2}+1\right)\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} \cdot x_{2}\right] \\
& =\left(m^{2}+1\right)\left(\frac{16 p^{2}}{m^{4}}+\frac{16 p^{2}}{m^{2}}\right) \\
& =\frac{16 p^{2}\left(m^{2}+1\right)^{2}}{m^{2}} .
\end{aligned}
$$

Therefore,
$$
|A B|=\frac{4 p\left(m^{2}+1\right)}{m^{2}}
$$

By substituting $m$ with $-\frac{1}{m}$, we get
$$
|C D|=\frac{4 p\left(\frac{1}{m^{2}}+1\right)}{\frac{1}{m^{2}}}=4 p\left(m^{2}+1\right) .
$$

Thus,
$$
\begin{aligned}
|A B|+|C D| & =\frac{4 p^{2}\left(m^{2}+1\right)}{m^{2}}+4 p\left(m^{2}+1\right)=4 p\left(m^{2}+1\right)\left(\frac{1}{m^{2}}+1\right) \\
& =4 p\left(m^{2}+\frac{1}{m^{2}}+2\right) \geqslant 4 p\left(2 \sqrt{m^{2} \cdot \frac{1}{m^{2}}}+2\right) \\
& =16 p .
\end{aligned}
$$

Equality holds when $m^{2}=\frac{1}{m^{2}}$, i.e., $m= \pm 1$. Therefore, when the lines $AB$ and $CD$ make angles of $45^{\circ}$ and $135^{\circ}$ with the $x$-axis, respectively, $|A B|+|C D|$ is minimized, and the minimum value is $16 p$.