The sum of all angles inside the triangle DBC must be 180. The sum of all angles inside the triangle AEC must be 180. The sum of all angles inside AEDB must be 360. All conditions together give the following matrix:
[tex] \left[\begin{array}{ccccccc}1&1&0&0&0&0&1\\0&0&0&0&1&1&1\\0&0&1&0&1&0&0\\0&0&0&1&0&1&0\\1&1&1&1&0&0&0\\1&0&0&1&0&0&0\\0&1&1&0&0&0&0\end{array}\right] + \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{array}\right] = \left[\begin{array}{c}180\\180\\180\\180\\360\\180\\180\end{array}\right] [/tex]
The solution shows, that the two triangles are similar and that EC/BC = AC/DC
15/6 = (2x² +7) / (x² + 1) = 2 + 5 /(x² + 1)
x² = 9
x = +/- 3
