Question

四边形ABCD中,对角线AC、BD相交于点M,且AC⊥AB,BD⊥CD,过点A作AE⊥BC,垂足为E,交BD于点F

Answer 1

MA·MC=MB·MD,则MA/MB=MD/MC;又∠AMD=∠BMC.则⊿AMD∽⊿BMC.

故:∠MAD=∠MBC; ∠MDA=∠MCB.

∠BEF=∠BDC=90度,∠EBF=∠DBC,则:⊿EBF∽⊿DBC,得BE/BF=BD/BC,BE·BC=BF·BD;

∠AEB=∠BAC=90度,∠ABE=∠CBA,则:⊿ABE∽⊿CBA,得AB/BE=BC/AB,BE·BC=AB².

∴AB²=BF·BD;又AD²=BF·BD.

则:AB=AD,∠ABD=∠ADB=∠MBC=∠BAE,得BF=AF.

设EF=X,则BF=AF=2-X.

BE²+EF²=BF²,即1+X²=(2-X)²,X=3/4.--------------------------即EF=3/4.