(1)设AP与DQ的交点为E
∵DQ⊥AP,∴PE为直角三角形ADP斜边AP上的高,∴∠PAD=∠CPQ,
又∵AD=DC,∠ADP=∠DCQ=90°,∴△ADP≌△DCQ,∴DP=CQ.
(2)由(1)知DP=CQ,又∵OC=OD,∠OCQ=∠ODP,∴△OCQ≌△ODP,∴∠COQ=∠DOP,在等式两边同时加上∠POC,即∠COQ+∠POC =∠DOP+∠POC,得∠POQ=∠DOC=90°,∴OP⊥OQ
(1)设AP与DQ的交点为E
∵DQ⊥AP,∴PE为直角三角形ADP斜边AP上的高,∴∠PAD=∠CPQ,
又∵AD=DC,∠ADP=∠DCQ=90°,∴△ADP≌△DCQ,∴DP=CQ.
(2)由(1)知DP=CQ,又∵OC=OD,∠OCQ=∠ODP,∴△OCQ≌△ODP,∴∠COQ=∠DOP,在等式两边同时加上∠POC,即∠COQ+∠POC =∠DOP+∠POC,得∠POQ=∠DOC=90°,∴OP⊥OQ