Phần III:

$\begin{array}{l}
III)\\
a)\dfrac{a}{b} = \dfrac{c}{d} = k \Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{a}{{a - b}} = \dfrac{{b.k}}{{b.k - b}} = \dfrac{{b.k}}{{b.\left( {k - 1} \right)}} = \dfrac{k}{{k - 1}}\\
\dfrac{c}{{c - d}} = \dfrac{{d.k}}{{d.k - d}} = \dfrac{{d.k}}{{d.\left( {k - 1} \right)}} = \dfrac{k}{{k - 1}}
\end{array} \right.\\
 \Rightarrow \dfrac{a}{{a - b}} = \dfrac{c}{{c - d}}\\
b)\dfrac{{a + c}}{{b + d}} = \dfrac{{b.k + d.k}}{{b + d}} = \dfrac{{k.\left( {b + d} \right)}}{{b + d}} = k\\
 \Rightarrow \dfrac{a}{b} = \dfrac{{a + c}}{{b + d}} = k\\
c)\dfrac{a}{{3a + b}} = \dfrac{{b.k}}{{3.b.k + b}} = \dfrac{{b.k}}{{b.\left( {3k + 1} \right)}} = \dfrac{k}{{3k + 1}}\\
\dfrac{c}{{3c + d}} = \dfrac{{d.k}}{{3.d.k + d}} = \dfrac{k}{{3k + 1}}\\
 \Rightarrow \dfrac{a}{{3a + b}} = \dfrac{c}{{3c + d}}\\
d)\dfrac{{a.c}}{{b.d}} = \dfrac{{b.k.d.k}}{{b.d}} = {k^2}\\
\dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}} = \dfrac{{{b^2}{k^2} + {d^2}{k^2}}}{{{b^2} + {d^2}}} = {k^2}\\
 \Rightarrow \dfrac{{a.c}}{{b.d}} = \dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}}\\
e)\dfrac{{a.b}}{{c.d}} = \dfrac{{b.b.k}}{{d.d.k}} = \dfrac{{{b^2}}}{{{d^2}}}\\
\dfrac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}} = \dfrac{{{b^2}.{k^2} - {b^2}}}{{{d^2}{k^2} - {d^2}}} = \dfrac{{{b^2}\left( {{k^2} - 1} \right)}}{{{d^2}\left( {{k^2} - 1} \right)}} = \dfrac{{{b^2}}}{{{d^2}}}\\
 \Rightarrow \dfrac{{a.b}}{{c.d}} = \dfrac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}}\\
f)\dfrac{{a.b}}{{c.d}} = \dfrac{{{b^2}}}{{{d^2}}}\\
\dfrac{{{{\left( {a - b} \right)}^2}}}{{{{\left( {c - d} \right)}^2}}} = \dfrac{{{b^2}{{\left( {k - 1} \right)}^2}}}{{{d^2}{{\left( {k - 1} \right)}^2}}} = \dfrac{{{b^2}}}{{{d^2}}}\\
 \Rightarrow \dfrac{{a.b}}{{c.d}} = \dfrac{{{{\left( {a - b} \right)}^2}}}{{{{\left( {c - d} \right)}^2}}}
\end{array}$

Phần IV câu 3:

Gọi chiều dài 3 cạnh là x;y;z (x;y;z>0)

Chu vi là 22cm nên x+y+z=22 (cm)

Vì 3 cạnh tỉ lệ với 2;4;5 nên ta có pt:

$\begin{array}{l}
\dfrac{x}{2} = \dfrac{y}{4} = \dfrac{z}{5} = \dfrac{{x + y + z}}{{2 + 4 + 5}} = \dfrac{{22}}{{11}} = 2\\
 \Rightarrow \left\{ \begin{array}{l}
x = 4\left( {cm} \right)\\
y = 8\left( {cm} \right)\\
z = 10\left( {cm} \right)
\end{array} \right.
\end{array}$

Vậy chiều dài 3 cạnh của hình tam giác là 4cm; 8cm; 10cm.