如图,正方形ABCD中,E点在边BC上,F点在边CD上,AF⊥ED. (1)线段AF和DE相等吗?说明理由; (2)求证:EF2=BE2+FD2.
问题描述:
如图,正方形ABCD中,E点在边BC上,F点在边CD上,AF⊥ED.
(1)线段AF和DE相等吗?说明理由;
(2)求证:EF2=BE2+FD2.
答
(1)AF=DE.∵四边形ABCD是正方形,∴AD=BC=CD,∠ADF=∠DCE=90°,∴∠DAF+∠DFA=90°∵AF⊥ED,∴∠DFA+∠EDC=90°,∴∠DAF=∠EDC,在△ADF和△DCE中,∵∠DAF=∠EDCAD=DC∠ADF=∠DCE,∴△ADF≌△DCE(ASA)...