Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
A\left( { - 1;2} \right),B\left( {3;2} \right),C\left( {1;4} \right)\\
\left\{ \begin{array}{l}
AB = \sqrt {{{\left( {3 - \left( { - 1} \right)} \right)}^2} + {{\left( {2 - 2} \right)}^2}}  = 4\\
AC = \sqrt {{{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left( {4 - 2} \right)}^2}}  = 2\sqrt 2 \\
BC = \sqrt {{{\left( {1 - 3} \right)}^2} + {{\left( {4 - 2} \right)}^2}}  = 2\sqrt 2 
\end{array} \right.\\
 \Rightarrow \Delta ABC;AC = BC;A{C^2} + B{C^2} = A{B^2}
\end{array}$

$ \Rightarrow \Delta ABC$ vuông cân ở $C$

Vậy $AC = BC = 2\sqrt 2 ;AB = 4$

b) Ta có:

$\Delta ABC$ vuông cân ở $C$

$\to \widehat{C}=90^0;\widehat{A}=\widehat{B}=45^0$

Vậy $\widehat{C}=90^0;\widehat{A}=\widehat{B}=45^0$

c) Ta có:

${S_{ABC}} = \dfrac{1}{2}AC.BC = \dfrac{1}{2}.2\sqrt 2 .2\sqrt 2  = 4$

Vậy ${S_{ABC}} = 4$

d) Ta có:

$\begin{array}{l}
AM = \sqrt {\dfrac{{2\left( {A{B^2} + A{C^2}} \right) - B{C^2}}}{4}}  = \sqrt {10} \\
BN = \sqrt {\dfrac{{2\left( {B{A^2} + B{C^2}} \right) - A{C^2}}}{4}}  = \sqrt {10} \\
CP = \sqrt {\dfrac{{2\left( {C{A^2} + C{B^2}} \right) - A{B^2}}}{4}}  = 2
\end{array}$

Vậy $AM = BN = \sqrt {10} ;CP = 2$

e) Ta có:

$\begin{array}{l}
R = \dfrac{{AB.AC.BC}}{{4{S_{ABC}}}} = \dfrac{{4.2\sqrt 2 .2\sqrt 2 }}{{4.4}} = 2\\
r = \dfrac{{{S_{ABC}}}}{{\left( {\dfrac{{AB + BC + AC}}{2}} \right)}} = \dfrac{4}{{\left( {\dfrac{{4 + 2\sqrt 2  + 2\sqrt 2 }}{2}} \right)}} = 2\sqrt 2  - 2
\end{array}$

Vậy $R = 2$ và $r = 2\sqrt 2  - 2$