AD=AB ∠DAQ=∠BAQ AQ=AQ
∴△ADQ≌△ABQ ······································· 2分
(2)解：△ADQ的面积恰好是正方形ABCD面积的
1时， 6过点Q作QE⊥AD于E,QF⊥AB于F，则QE = QF
118AD?QE=S正方形ABCD= 2634 ∴QE= ··························································································· 4分
3QEDE?由△DEQ ∽△DAP得 解得AP?2 APDA1∴AP?2时，△ADQ的面积是正方形ABCD面积的 ··························· 5分
6（3）若△ADQ是等腰三角形，则有 QD=QA或DA=DQ或AQ=AD ①当点P运动到与点B重合时，由四边形ABCD是正方形知 QD=QA 此时△ADQ是等腰三角形
②当点P与点C重合时，点Q与点C也重合，
此时DA=DQ, △ADQ是等腰三角形 ③解：如图，设点P在BC边上运动到CP?x时，有AD=AQ ∵ AD∥BC ∴∠ADQ=∠CPQ 又∵∠AQD=∠CQP ∠ADQ=∠AQD ∴∠CQP=∠CPQ ∴ CQ=CP=x
∵AC=42 AQ = AD =4 ∴x?CQ?AC?AQ?42?4
即当CP?42?4时，△ADQ是等腰三角形 ······························ 8分
