It is known that AB and CD intersect on O, m and N on AB, and AC = BD, am = BN, DM = CN

It is known that AB and CD intersect on O, m and N on AB, and AC = BD, am = BN, DM = CN

∵ AM=BN ∴AN=BM
AC = BD CN = DM
≌ triangle ACN ≌ triangle BDM
∴∠A=∠B
∵AC=BD ∠AOC=∠BOD
≌ triangle AOC ≌ triangle BOD
∴OA=OB,OC=OD
The AB and CD are equally divided

In the triangle ABC, the angle ACB is 90 degrees, CM is perpendicular to AB and m, at is the bisector of the angle BAC, intersecting cm at D, passing point D as De, parallel to AB, intersecting BC at E
Verification: CT = be

Make FD ‖ BC cross AB to F
∵CM⊥AM BC⊥AC
∴∠ACD=∠ABC=∠AFD
Ad = ad
∴△AFD≌△ACD
∴FD=CD
∵ at bisection ∠ BAC ∠ amd = 90 ° = ∠ act
∴∠ATC=∠ADM=∠CDT
∴CT=CD=FD
FD ‖ be de ‖ FB
The BEDF is a parallelogram
∴BE=DF
∴CT=BE

In triangle ABC, points D and E are on AC and ab respectively, BM bisector angle ABC, CM bisector angle ACB, en bisector angle bed and DN bisector angle EDC
The faster, the more points
To prove that a, m and N are on the same straight line

This is a wrong problem without proof. It is impossible for a, B and C to be in the same line