?AH?5 2(3)如图,延长FE、BC,交于点M,作CN?EF于点N
3QBF?6,cos?FBG?
5?cos?FBC?BF3? BM5?63? BM5?BM?10
?MF?8 QBG?5
∴点G为BM中点 ∴点N为AF的中点,
?NG?11BF?3?NF?MF?4 22Q?ENG??DEG??DFE?90?
??NEG??NGE?90?,?NEG??FED?90?
??NGE??FED
?VENG:VDFE,
?NGEN ?EFDF设EF?a
?34?a ?aDF14?DF?a(4?a)?
33解得a?2
?EF?2.
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