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Fengxian Gao San mo er mathematics
Proof: (1)∵AD⊥AC, AE⊥AB,

∴∠DAC=∠BAE=90,

∴∠DAB=∠EAC,

AD = BD,AE=EC,

∴∠DAB=∠DBA,∠ECA=∠EAC,

∴∠DBA=∠ECA,

∴△ADB∽△AEC,

∴S△ADBS△AEC=(ABAC)2,

∫∠BCA = 90 degrees,

cos∠BAC=45=ACAB,

∴s△adbs△aec=25 16;

(2) The intersection point E is EH⊥AC, and AB extends to G to connect DG.

AE = EC,

∴AH=CH,EH⊥AC,

∫∠BCA = 90 degrees,

∴GH∥BC,

∴AG=BG,

AD = BD,

∴DG⊥AB,

∵AD⊥AC,AE⊥AB,

∴GE∥AD,DG∥AE,

∴ Quadrilateral is a parallelogram,

∴AF=GF,

∴AFFB= 13.