Analysis: parallel lines passing through P are BC, and passing through AC is M; Then △APM is also an equilateral triangle. In the equilateral triangle APM, PE is the height on AM, and AE = EM according to the unity of three lines of equilateral triangle. It is easy to get △ PMD △ QCD, then DM = CD. At this time, it is found that the length of DE is exactly half that of AC, and it is solved. Solution: p is PM‖BC, and AC is in m;

It is easy to know that △APM is an equilateral triangle;

And ∵PE⊥AM,

∴ae=em； (equilateral triangle with three lines in one)

‖CQ in the afternoon,

∴∠pmd=∠qcd,∠mpd=∠q；

∫PA = PM = CQ，

∴△pmd≌△qcd；

∴cd=dm； DM+EM = AE+CD = 1/2ac = 1/2。