The problem of proving the first volume of mathematics in grade three.

Do ME//AB to m and cross AC to e.

Obviously ME= 1/2 AB (triangle midline theorem)

E is the midpoint of communication.

Angle EMC= angle B.

AD is perpendicular to BC

Therefore, in a right-angled triangular ADC, the connection line DE is obviously the center line of its hypotenuse AC.

So DE= 1/2 AC = CE = AE.

Therefore, the triangle CDE is an isosceles triangle.

Angle CDE= angle c

The other angle EMC is the outer angle of the triangle EDM.

So angle EMC= angle EDM+ angle DEM.

Also: Angle B = 2° Angle C.

Angle B= angle EMC

Angle EDM= angle c

So angle DEM= angle EDM= angle C.

Therefore, the triangle DEM is an isosceles triangle,

DM=EM

And EM= 1/2 AB.

So DM= 1/2 AB.