Question research 20 15 mathematical answers

For the research on (1), see the analysis and proof.

(2)① See analysis and verification.

② 。

Analysis of test questions: (1) Application ideas 1: BM=CM=MA can be obtained according to the conditions, and ∠ 1=∠B, ∠2=∠C can be obtained from the properties of isosceles triangles, and the conclusion can be drawn from the sum theorem of the internal angles of triangles.

(2)① Connecting OD and CD, we can get AO=OD=OC=a from the nature of the circle, then we can get that △ODC is an equilateral triangle from the conditions, and we can get ∠ BDC = 30 from the relationship between the outer angle and the inner angle, so we can get ∠ ODB = 90 and draw a conclusion.

② Using the conclusion of (1), we can get ∠ ADB = ∠ ACE = 90, from which we can get △ADB∽△AEC, from the similarity property we can get △ADE∽△ABC, from the area ratio of similar triangles is equal to the square of the similarity ratio, and finally we can get the conclusion from the function value of the acute triangle.

Solution: (1) problem research, application ideas 1:

∵M is the midpoint of∴ BM = cm = in 500 BC.

∫ Ma = BC,∴BM=CM=MA.

∴∠ 1=∠B,∠2=∠C。

≈ 1+≈b+≈2+≈c = 180 ,∴2∠ 1+2∠2= 180 .

∴∠ 1+∠ 2 = 90, that is ∠ BAC = 90.

(2)① Proof: Connect OD, CD,

∫∠DAB = 30，OA=a，

∴AO=OD=OC=a,∠BOD=2∠A=60 .

△ ODC is an equilateral triangle.

∴CD=OC=a,∠DCO=∠CDO=60 .

∵OB=2a,∴BC=a。 ∴BC=DC。 ∴∠B=∠BDC。