As shown in the figure, D is a point on the side BC of △ ABC, and ∠ 1 = 2, ∠ 3 = 4, ∠ BAC = 63 ° to find the degree of ∠ DAC?

As shown in the figure, D is a point on the side BC of △ ABC, and ∠ 1 = 2, ∠ 3 = 4, ∠ BAC = 63 ° to find the degree of ∠ DAC?

In this way, ∠ 3 = 1 + 2, ∠ 1 = 2, so ∠ 3 = 2 * 2
63-∠2=180-∠3-∠4
63-∠2=180-4*∠2,∠2=39,∠DAC=63-39=24
∠DAC=24°!

As shown in the figure, △ ABC, ad ⊥ BC is at point D, ad = DC, point F is on ad, ∠ FCD = ∠ BAD.BF The intersection of extension line AC and point e
(1) Verification: be ⊥ AC
(2) Let the length of CE be m and AC = BF be expressed by an algebraic expression containing M

(1) It is proved that ∵ ad ⊥ BC is at point D,
∴∠ADB=∠ADC=90°,
In △ abd and △ CFD
∠BAD＝∠FCD    
AD＝DC   
∠ADB＝∠CDF    
∴△ABD≌△CFD（ASA）,
∴BD=DF,
∴∠FBD=∠BFD=45°,
∴∠AFE=∠BFD=45°,
And ∵ ad = DC,
∴∠DAC=∠ACD=45°,
∴∠AEF=90°,
∴BE⊥AC．
∵∠EBC=∠ACD=45°,CE=m,
∴BE=CE=m
And ∵ ∠ AFE = ∠ FAE = 45 °,
∴AE=FE,
∴AC+BF
=CE+AE+BF
=CE+EF+BF
=CE+BE
=CE+CE
=2m．

It is known that: as shown in the figure, △ ABC, ad ⊥ BC is at point D, ad = DC, ∠ FCD = ∠ bad, point F is on ad, and the extension line of BF intersects AC at point E. (1) prove: be ⊥ AC; (2) let the length of CE be m, and use the algebraic formula containing m to express AC + BF

(1) It is proved that ∵ ad ⊥ BC is located at point D, ≌ ADC = 90 ° in △ abd and △ CFD, ≌ BD = DF, ≌ FBD = BFD = 45 ° and ≌ AFE = BFD = 45 ° and ≌ ad = DC, ≌ DAC = ACD = 45 ° and ≌ a