29. As shown in the figure, in △ABC, ∠ ACB > ∠ ABC, write ∠ACB-∠ABC=α, AD is the bisector of △ABC, M is a point on DC, ME is perpendicular to the straight line where AD is, and E is vertical.

(1) expresses the value of ∠DME with the algebraic expression of α;

(2) If point M moves on ray BC (not coincident with point D), and other conditions remain unchanged, does the size of ∠DME change with the position of point M? Please draw a picture, give your conclusion and explain the reasons.

Solution: (1) Solution 1: Make a straight line EM intersect AB at point F, and the extension line intersect AC at point G (see figure 1).

∫AD splitting∠ ∠BAC,

∴∠ 1 = ∠ 2.( 1)

∵ME⊥AD，

∴∠AEF=∠AEG=90

∴∠3=∠G.

∠∠3 =∠b+∠DME，

∴∠ACB=∠G+∠GMC=∠G+∠DME，

∴∠B+∠DME=∠ACB-∠DME.

∴∠dme= 1 2(≈ACB-∠b)=α2； (2 points)

Scheme 2: As shown in Figure 2 (without auxiliary lines),

∫AD splitting∠ ∠BAC,

∴∠ 1 = ∠ 2.( 1)

∵ME⊥AD，

∴∠DEM=90，∠ADC+∠DME=90。

∠∠ADB =∠2+∠C = 90+∠DME，

∴∠DME=∠2+∠C-90。

∫∠ADC =∠ 1+∠B，

∴∠ 1=∠ADC-∠B.

∴∠dme=∠ 1+∠c-90 =(∠ADC-∠b)+∠c-90

=∠C-∠B-(90 -∠ADC)=∠C-∠B-∠DME

∴∠dme= 1 2(∠c-∠b)=α2； (2 points)

(2) As shown in Figures 3 and 4, the size of ∠DME remains unchanged when point M moves on the ray BC (not coincident with point D) (the same is true when point M moves to points B and C).

Prove 1: set point m moves to m', and make m' e' ⊥ ad at point e' through point m'

∵m′e′⊥ad，

∴me∥m′e′.

∴∠ DM' e' = ∠ DME = α 2。 (4 points)

By the way, he said that the map I copied was illegal, and there was nothing I could do.