It is known that point C is a point on the line segment AB, with AC and BC as sides, △ACD and △BCE as the same side of the line segment AB, CA=CD, CB=CE, ∠ACD=∠BCE, and the straight line AE and BD intersect at point F,

(1) as shown in figure 1, if ∠ ACD = 60, ∠ AFB =120; As shown in Figure 2, if ∠ ACD = 90, ∠ AFB = 90; As shown in Figure 3, if ∠ ACD = 120 and ∠ AFB = 60;

(2) As shown in Figure 4, if ∠ACD=α, ∠ AFB = 180-α (expressed by a formula containing α). Test center: congruent triangles's judgment; The nature of congruent triangles. Analysis: (1) As shown in figure 1, firstly, it is proved that △ BCD △ ECA, and ∠EAC=∠BDC, and then its degree is obtained according to the external angle that ∠AFB is △ADF.

As shown in Figure 2, firstly, it is proved that △ ace △ DCB, then ∠AEC=∠DBC, and ∠FDE=∠CDB, and then ∠ AFB = 90.

As shown in Figure 3, △ ace △ DCB is first proved, and then ∠EAC=∠BDC+∠ FBA = 180-∠ DCB gets ∠ FAB+∠ FBA.

(2) Get ∠ACE=∠DCB from ∠ACD=∠BCE, and then get ∠CAE=∠CDB from the triangle interior angle sum theorem, so as to get ∠DFA=∠ACD and get the conclusion ∠.

So △ACD is an equilateral triangle

CB = CE，∠ACD=∠BCE=60

So △ECB is an equilateral triangle

∫AC = DC，∠ACE=∠ACD+∠DCE，∠BCD=∠BCE+∠DCE

Also material ≈ACD =∠BCE

∴∠ACE=∠BCD

∫AC = DC，CE=BC

∴△ACE≌△DCB

∴∠EAC=∠BDC

∠AFB is the external angle of△△ ADF.

∴∠afb=∠adf+∠fad=∠adc+∠cdb+∠fad=∠adc+∠eac+∠fad=∠adc+∠dac= 120

As shown in figure 2, AC = CD, ∠ ACE = ∠ DCB = 90, EC=CB.

∴△ACE≌△DCB

∴∠AEC=∠DBC，

∠∠FDE =∠CDB，∠ DCB = 90。

∴∠EFD=90

∴∠AFB=90

As shown in Figure 3, ∫∠ACD =∠BCE.

∴∠ACD-∠DCE=∠BCE-∠DCE

∴∠ACE=∠DCB

And ∵CA=CD, CE=CB.

∴△ACE≌△DCB

∴∠EAC=∠BDC

∠∠BDC+∠FBA = 180-∠DCB = 180-( 180-∠ACD)= 120

∴∠FAB+∠FBA= 120

∴∠AFB=60

So fill in120,90,60.

(2)∫∠ACD =∠BCE

∴∠ACD+∠DCE=∠BCE+∠DCE

∴∠ACE=∠DCB

∴∠CAE=∠CDB

∴∠DFA=∠ACD

∴∠AFB = 180-∠DFA = 180-∠ACD = 180-α。 Comments: This topic examines congruent triangles's judgment and its nature, triangle interior angle and theorem.

Turn left | turn right