Set the coordinate system: the origin O is at point A, the Y axis is straight, the point B is in the first quadrant, and the point C is in the second quadrant. ?

Suppose angle BAT= angle CAT=P, AB=c, AC=b, BD=CE=d?

So: angle BAX=pai/2? -P； Angle CAX=pai/2? +P？

B point coordinates: B[c*cos(pai/2? -P)、c*sin(pai/2？ -P)]=B[c*sinP，c*cosP]？

C point coordinates: C[b*cos(pai/2? +P)、b*sin(pai/2？ +P)]=C[-b*sinP，b*cosP]？

D point coordinates: D[(c-d)cos(pai/2? -P)，(c-b)sin(pai/2？ -P)]=D[(c-d)sinP，(c-b)cosP]？

Point e coordinates: E[(c-d)cos(pai/2? +P)，(c-b)sin(pai/2？ +P)]=E[-(c-d)sinP，(c-b)cosP]？

Therefore, the coordinates of point m: m [(c * SINP-b * SINP)/2, (c * COSP+b * COSP)/2]?

Point n coordinates: n {[(c-d) SINP-(c-d) SINP]/2, [(c-b) COSP+(c-b) COSP]/2}?

=? N{(c*sinP-b*sinP)/2，[(c-b)cosP+(c-b)cosP]/2}？

The abscissas of point m and point n are equal. So: MN||AT?

Complete the certificate. ?

(2) Proof 2:? Please look at the pictures.

Complete the certificate.