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\begin{center}
    \begin{tikzpicture}
        \def\a{1.5};
        \def\b{2};
        \def\c{1};

\def\x{1.5};

\coordinate (A) at (-\a, 0);
        \coordinate (C) at (\c, -\b);
        \coordinate (X) at (0, -\x);

\node at (A) [left] {woman};
        \node at (C) [right] {child};

\filldraw (A) circle (1pt);
        \filldraw (C) circle (1pt);

\draw[dashed] (A) -- (X) -- (C);

\node at ({-\a/2}, 0) [above] {$a$};
        \draw[->] ({-\a/2}, 0) -- (A);
        \draw[->] ({-\a/2}, 0) -- (0,0);

\node at  (-0.1, {-(\x+\b)/2})  [left] {$x$};
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\b); 
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\x);

\node at (\c, {-\b/2}) [right] {$b$};
        \draw[->] (\c, {-\b/2}) -- (\c, 0); 
        \draw[->] (\c, {-\b/2}) -- (\c, -\b);

\node at ({\c/2}, -\b) [below] {$c$};
        \draw[->] ({\c/2}, -\b) -- (C);
        \draw[->] ({\c/2}, -\b) -- (0, -\b);

\draw (0,1) -- (0,-3);

\filldraw[color=blue, opacity=0.1] (0,1) -- (0, -3) -- (2, -3) -- (2, 1);
        \filldraw[color=green, opacity=0.1] (0,1) -- (0, -3) -- (-3, -3) -- (-3, 1);
        
    \end{tikzpicture}
\end{center}

The distance to where she enters the water is $\sqrt{a^2+(b-x)^2}$ and the distance through the water is $\sqrt{x^2+c^2}$.

The total time will be $\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}$.

To minimise this, we can differentiate.

\begin{align*}
\frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\
\Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12}
\end{align*}

When she is in the water, she can will move with velocity $\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}$. She needs to travel a distance $\begin{pmatrix} c \\ -x \end{pmatrix}$, so we must have that 
\begin{align*}
&& \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\
\Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\
\Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\
\Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\
\Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\
\Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2}  \\
&& \cos \theta &=  \frac{2xc}{c^2+x^2}  \\
\end{align*}
(where we have taken the positive value for $\cos \theta$ since we must be heading towards the child).

Since $v \cos \theta t = c$ the time taken to reach the child in the water is $\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}$.

So the total time is: $\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}$.

To minimise this, we can differentiate.

\begin{align*}
\frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\
\Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x) 
\end{align*}

as required.

When $b = c$, the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be:

\begin{center}
    \begin{tikzpicture}
        \def\a{1.5};
        \def\b{1.5};
        \def\c{1.5};

\def\x{0};

\coordinate (A) at (-\a, 0);
        \coordinate (C) at (\c, -\b);
        \coordinate (X) at (0, -\x);

\node at (A) [left] {woman};
        \node at (C) [right] {child};

\filldraw (A) circle (1pt);
        \filldraw (C) circle (1pt);

\draw[dashed] (A) -- (X) -- (C);

\node at ({-\a/2}, 0) [above] {$a$};
        \draw[->] ({-\a/2}, 0) -- (A);
        \draw[->] ({-\a/2}, 0) -- (0,0);

\node at  (-0.1, {-(\x+\b)/2})  [left] {$x$};
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\b); 
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\x);

\node at (\c, {-\b/2}) [right] {$b$};
        \draw[->] (\c, {-\b/2}) -- (\c, 0); 
        \draw[->] (\c, {-\b/2}) -- (\c, -\b);

\node at ({\c/2}, -\b) [below] {$c$};
        \draw[->] ({\c/2}, -\b) -- (C);
        \draw[->] ({\c/2}, -\b) -- (0, -\b);

\draw (0,1) -- (0,-3);

If $b > c$ then she should run a little downstream first.

\begin{center}
    \begin{tikzpicture}
        \def\a{1.5};
        \def\b{2};
        \def\c{1.5};

\def\x{0.25};

\coordinate (A) at (-\a, 0);
        \coordinate (C) at (\c, -\b);
        \coordinate (X) at (0, -\x);

\node at (A) [left] {woman};
        \node at (C) [right] {child};

\filldraw (A) circle (1pt);
        \filldraw (C) circle (1pt);

\draw[dashed] (A) -- (X) -- (C);

\node at ({-\a/2}, 0) [above] {$a$};
        \draw[->] ({-\a/2}, 0) -- (A);
        \draw[->] ({-\a/2}, 0) -- (0,0);

\node at  (-0.1, {-(\x+\b)/2})  [left] {$x$};
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\b); 
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\x);

\node at (\c, {-\b/2}) [right] {$b$};
        \draw[->] (\c, {-\b/2}) -- (\c, 0); 
        \draw[->] (\c, {-\b/2}) -- (\c, -\b);

\node at ({\c/2}, -\b) [below] {$c$};
        \draw[->] ({\c/2}, -\b) -- (C);
        \draw[->] ({\c/2}, -\b) -- (0, -\b);

\draw (0,1) -- (0,-3);

and if $c > b$ she should actually run a little upstream to take advantage of the current:

\begin{center}
    \begin{tikzpicture}
        \def\a{1.5};
        \def\b{1.5};
        \def\c{2};

\def\x{-0.25};

\coordinate (A) at (-\a, 0);
        \coordinate (C) at (\c, -\b);
        \coordinate (X) at (0, -\x);

\node at (A) [left] {woman};
        \node at (C) [right] {child};

\filldraw (A) circle (1pt);
        \filldraw (C) circle (1pt);

\draw[dashed] (A) -- (X) -- (C);

\node at ({-\a/2}, 0) [above] {$a$};
        \draw[->] ({-\a/2}, 0) -- (A);
        \draw[->] ({-\a/2}, 0) -- (0,0);

\node at  (-0.1, {-(\x+\b)/2})  [left] {$x$};
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\b); 
        \draw[->] (-0.1, {-(\x+\b)/2}) -- (-0.1, -\x);

\node at (\c, {-\b/2}) [right] {$b$};
        \draw[->] (\c, {-\b/2}) -- (\c, 0); 
        \draw[->] (\c, {-\b/2}) -- (\c, -\b);

\node at ({\c/2}, -\b) [below] {$c$};
        \draw[->] ({\c/2}, -\b) -- (C);
        \draw[->] ({\c/2}, -\b) -- (0, -\b);

\draw (0,1) -- (0,-3);

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