设角A的对边为a,角B的对边为b,角C的对边为c.则由正弦定理：a/sinA=b/sinB=c/sinC=2R（R为外接圆半径）由已知条件得：（a/2R)^2+(b/2R)^2=(c/2R)^2,展开并同时乘以（2R)^2得到：a^2+b^2=c^2,所以三角形ABC是直角三角...