So the triangle AEC is an isosceles triangle,

Because d is a point on that vertical line in BC,

So DB=DC,

So the triangle DBC is also an isosceles triangle,

Because ∠ACE=∠B,

So delta δ△AEC is similar to delta δ△DBC,

So EC/AC=DC/BC,

And because ∠DCB=∠B=∠ACE,

So ∠ACE+∠DCA=∠DCB+∠DCA,

∠DCE=∠BCA，

So delta △DCE is similar to delta △ABC,

So ∠EDC=∠B, and ∠B=∠DCB,

So ∠EDC=∠DCB,

So DE//BC