C is a point on the line AB, with AC and BC as edges, make equilateral △ ACD and equilateral △ BCE on the same side of AB, AE and CD intersect at m, BD and CE intersect at n It is proved that (1) △ MCN is an equilateral triangle. (2) if AC: CB = 2:1, then de ⊥ CE

C is a point on the line AB, with AC and BC as edges, make equilateral △ ACD and equilateral △ BCE on the same side of AB, AE and CD intersect at m, BD and CE intersect at n It is proved that (1) △ MCN is an equilateral triangle. (2) if AC: CB = 2:1, then de ⊥ CE

one
AC=DC
∠ACE=∠DCB
CE=CB
So Δ ace ≌ Δ DCB
∠CAE=∠CDB
AC=DC
∠ACM=∠DCN
So Δ ACM ≌ Δ DCN
So cm = CN
two
Connect De, take DC midpoint P, connect PE
Then PD = PC = DC / 2 = CD
∠DCE=60°
So Δ CPE is a positive Δ CPE
∠CEP=∠CPE=60°
∠PED=∠PDE=∠CPE/2=30°
Therefore, CED = 30 ° + 60 ° = 90 °
So de ⊥ CE
If the explanation is not clear enough,