[gools]

ข้อ 17 ครับ
\[\begin{array}{rcl}\because\qquad AB^{2}-BC^{2} & = & BD^{2}-AB^{2}=BE^{2}-BD^{2} \\
\text{จะได้ว่า }\qquad AB^{2} & = & \frac{BC^{2}+BD^{2}}{2} \\
\text{และ }\qquad BD^{2} & = & \frac{AB^{2}+BE^{2}}{2} \\
\therefore\qquad\qquad\quad\ \ AB^{2} & = & \frac{\frac{BE^{2}+AB^{2}}{2}+BC^{2}}{2} \\
3AB^{2}-2BC^{2} & = & BE^{2} \\
\therefore\qquad BE^{2}-AB^{2} & = & 3AB^{2}-2BC^{2}-AB^{2} \\
&=& 2(AB^{2}-BC^{2}) \\
&=& 4
\end{array}
\]