sossssssssssssssss

Ta có: 

$\begin{array}{l}
{S_{ABC}} = \dfrac{{abAC}}{{4R}} \Rightarrow ab = \dfrac{{4R.{S_{ABC}}}}{{AC}}\\
TT:\left\{ \begin{array}{l}
cd = \dfrac{{4R{S_{ADC}}}}{{AC}}\\
ad = \dfrac{{4R{S_{ABD}}}}{{BD}}\\
bc = \dfrac{{4R{S_{BCD}}}}{{BD}}
\end{array} \right.
\end{array}$

$\begin{array}{l}
T = \dfrac{{\left( {ab + cd} \right)\left( {ad + bc} \right)}}{{4S}}\\
 = \dfrac{{\left( {\dfrac{{{S_{ABC}}.4R}}{{AC}} + \dfrac{{{S_{ADC}}.4R}}{{AC}}} \right)\left( {\dfrac{{{S_{ABD}}.4R}}{{BD}} + \dfrac{{{S_{BCD}}.4R}}{{BD}}} \right)}}{{4S}}\\
 = \dfrac{{16{R^2}\left( {{S_{ABC}} + {S_{ADC}}} \right)\left( {{S_{ABD}} + {S_{BCD}}} \right)}}{{4S.AC.BD}}\\
 = \dfrac{{4{R^2}\left( {{S_{ABC}}.{S_{ABD}} + {S_{ADC}}.{S_{ABD}} + {S_{ABC}}.{S_{BCD}} + {S_{ADC}}.{S_{BCD}}} \right)}}{{S.AC.BD}}\\
 = \dfrac{{4{R^2}\left[ {{S_{ABC}}\left( {{S_{ABD}} + {S_{BCD}}} \right) + {S_{ADC}}\left( {{S_{ABD}} + {S_{BCD}}} \right)} \right]}}{{S.AC.BD}}\\
 = \dfrac{{4{R^2}\left[ {{S_{ABC}}.S + {S_{ADC}}.S} \right]}}{{S.AC.BD}} = \dfrac{{4{R^2}.{S^2}}}{{S.AC.BD}} = \dfrac{{4{R^2}S}}{{AC.BD}} = \dfrac{{4{R^2}S}}{{2S}} = \dfrac{{4{R^2}}}{2} = 2{R^2} = 2
\end{array}$

(Diện tích hình cánh diều $S=\dfrac{AC.BD}{2}$ )