如图,△ABC是等边三角形.AE=CD,AD,BE相交于点P.BQ⊥AD于Q,PQ=3,PE=1,求AD的长,

∵△ABC为等边三角形,∴AB=CA,∠BAE=∠ACD=60°；又∵AE=CD,∴△ABE≌△CAD；∴BE=AD,∠CAD=∠ABE；∴∠BPQ=∠ABE+∠BAD=∠BAD+∠CAD=∠BAE=60°；∵BQ⊥AD,∴∠AQB=90°,则∠PBQ=90°-60°=30°；∵PQ=3,∴在Rt△BPQ中,BP=2PQ=6；又∵PE=1,∴AD=BE=BP+PE=7．