Traditional Culture Encyclopedia - Photography major - (20 10? Suizhou) As shown in the figure, a point P is on the side AB of equilateral △ABC with a side length of 1, PE⊥AC is on E, and Q is on the extension line of BC. When PA=CQ,

(20 10? Suizhou) As shown in the figure, a point P is on the side AB of equilateral △ABC with a side length of 1, PE⊥AC is on E, and Q is on the extension line of BC. When PA=CQ,

Solution: p is PM∑BC, and AC is in m;

∫△ABC is an equilateral triangle, PM∨BC,

△ APM is an equilateral triangle;

And ∵PE⊥AM,

∴ae=em= 12am； (equilateral triangle with three lines in one)

∫PM∨CQ，

∴∠pmd=∠qcd,∠mpd=∠q；

∫PA = PM = CQ，

At △PMD and △QCD.

∠PDM=∠CDQ∠PMD=∠DCQPM=CQ

∴△pmd≌△qcd(aas)；

∴cd=dm= 12cm；

∴ de = DM+me =12 (am+MC) =12ac =12, so B.