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(2)如图2中,延长BA交CD的延长线于K.∵BN⊥CK,∴∠BNK=∠BNC=90°,∵BN=BN,∠NBK=∠NBC,∴△BNK≌△BNC(ASA),∴NK=NC,∵DN=1/3CN,∴DK=2DN,DK:KC=1:3,∵AD∥BC,∴△KAD∽△KBC,∴(S_(△KAD))/(S_(△KBC))=((KD)/(KC))2=1/9,∵S四边形ABCD=(32)/7,∴S△KAD=4/7,∴...
integrity sha512-zrsUxjLOKAzdewIDRWy9nsV1GQsKBCWaGwsZQlCgr6/q+vjyZhFgqedLfFBuI9anTPEUT4APq9Mu0SZBTzIcGQ== "@rollup/pluginutils@^4.1.0": version "4.1.2" resolved "https://registry.npmmirror.com/@rollup/pluginutils/download/@rollup/pluginutils-4.1.2.tgz#ed5821c15e5e05e32816f5fb9ec...
25.【思路分析】(1)根据两点之间线段最短解决问题即可;(2)如解图②,延长BA交CD的延长线于点K.利用相似三角形的性质求出S△ADK,再根据 S_(△KBN)=1/2S_(△KBC) 即可解决问题;(3)存在.如解图③,延长BA交CD的延长线于点E,连接AC,BD交于点P,此时PA+PB+PC+PD的值最小.在CA的延长线上取一点F,...
【解析】(1)如图1中,点P即为所求∫_(-π/2)^0=∫_0^1 E_XBA_CCDE (2)如图2中,延的延长线于K.NCB图2BN⊥CK∴.∠BNK=∠BNC=90°,BN=BN,∠NBK=∠NBC∴.△BNK≡△BNC(ASA)NK=NC,∵DN=1CN∴DK=2DN,DK:KC=1:3,∵AD|‖BC∴△KAD-△KBC(△KAD)/△=((KD)/(KC))^2=1/9 32S四...