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,
∵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.
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