∵BK⊥DK,∴∠BKD=∠BKF=90,
和∵被平分∠DBC,
∴∠DBK=∠FBK,
且∵BK=BK,
在△BKD和△BKF,
∠BKD=∠BKF=90 ∠DBK=∠FBK?BK=BK,
∴△BKD≌△BKF,
∴DK=FK,
∠∠BCE =∠DCF = 90,∠ CBE+∠ BEC = 90,∠ DEK+∠ EDK = 90,
並且∠BEC=∠DEK,
∴∠CBE=∠CDF,
∴△BEC∽△DCF,
∴DFBE=DCBC,
∫tan∠DBC = DCBC = 43,
∴be=34df=34×2df=32df;
(2)設DL=5a?那麽KL = 3a,
∴DK=8a,
∴BE=32DK=12a,
∠EDK=∠EBC=∠DBE,
∴tan∠EDK=∠DBK,
∴DKBK=EKDK,也就是EK8a=8aEK+12a
∴EK=4a,EK=12a(略),
∴DE=45a,
並且譚∠EBC =譚∠EDK,
∴ECBE=EKDK=4a8a=12,
∴BC=2EC,
∴EC=1255a,BC=2455a,
∴CD=3255a,BD=85a,
∠∠BED =∠IEL,
∴∠BEI=∠DEL,
∠∠IBE =∠LDE,
∴△DLE∽△LDE,
∴BIDL=BEDE,
∴BI5a=12a45a,
∴BI=35a,
∴DI=55a,
∴DIDB=58=DLDK,
∫△IDL?△BDK,
∴△DIL∽△DBK,
∴ILBE=DIDK=58,
∴BE=8?即16a=8,
∴a=12,
∴BI=35a=325.