Fortune Telling Collection - Free divination - As shown in figure 1, BD is the diagonal of rectangular ABCD, the bisector of ∠DBC intersects with DC at point E, and the extension line of DK ⊥ BE intersects with BE at K. (1) If tan∠DBC=43,

As shown in figure 1, BD is the diagonal of rectangular ABCD, the bisector of ∠DBC intersects with DC at point E, and the extension line of DK ⊥ BE intersects with BE at K. (1) If tan∠DBC=43,

(1) extends the intersection of DK and BC at point f,

∵BK⊥DK,∴∠BKD=∠BKF=90,

And ∵ divided equally ∠DBC,

∴∠DBK=∠FBK,

And ∵BK=BK,

At △BKD and △BKF,

∠BKD=∠BKF=90 ∠DBK=∠FBK? BK=BK,

∴△BKD≌△BKF,

∴DK=FK,

∠∠BCE =∠DCF = 90,∠ CBE+∠ BEC = 90,∠ DEK+∠ EDK = 90,

And ∠BEC=∠DEK,

∴∠CBE=∠CDF,

∴△BEC∽△DCF,

∴DFBE=DCBC,

∫tan∠DBC = DCBC = 43,

∴be=34df=34×2df=32df;

(2) Let DL=5a? Then KL = 3a,

∴DK=8a,

∴BE=32DK= 12a,

∠EDK=∠EBC=∠DBE,

∴tan∠EDK=∠DBK,

∴DKBK=EKDK, which means EK8a=8aEK+ 12a.

∴EK=4a, EK= 12a (omitted),

∴DE=45a,

And tan ∠EBC = tan ∠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? That is, 16a=8,

∴a= 12,

∴BI=35a=325.