>>930
いいえ。

外接円の半径を R とすると
　AB = 2R sin(∠AOB/2),
　BC = 2R sin(∠BOC/2),
　CD = 2R sin(∠COD/2),
　DA = 2R sin(∠DOA/2),
また題意より
　R = 65/8,
　BC = CD = 13,
　AB+BC+CD+DA = 44,
したがって
　AB + DA = 44 -13 -13 = 18,
　∠AOB/2 = 2arctan(4/7) = arcsin(56/65) = 59.4897626ﾟ
　∠BOC/2 = 2arctan(1/2) = arcsin(4/5) = 53.130102ﾟ
　∠COD/2 = 2arctan(1/2) = arcsin(4/5) = 53.130102ﾟ
　∠DOA/2 = 2arctan(1/8) = arcsin(16/65) = 14.25003ﾟ
よって
　AB = 14
　DA = 4