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The answer of the seventh grade math master
With a as the center, AE as the radius, BC intersecting with D, connecting ED.

AE=AD

∠∠CAB =∠CBA = 80,∠CAF=20,∠CBE=30

∴∠AEB=50,∠ABE=50

∴AB=AE=AD, ∠ ADB = 80 △ Abd is an isosceles triangle.

∴∠BAD=20

∴∠DAF=40,∠EAD=60

∴△AED is an equilateral triangle,

∠EDA=∠EAD=∠AED=60

∴∠daf=40 =∠ Air base,

△ ADF is an isosceles triangle,

∠∠ADB = 80,∠EDA=60

∴∠EDF=40

∴∠EFD=∠FED=70

∴∠EFA=∠EFD-∠AFB=70 -40 =30