Mathematical geometry problems in grade three

1, BE is the angular bisector, and BE⊥AD,BE=BE.

Triangle ABE is equal to triangle DBE.

So e is the midpoint.

2. take CG=CA.

Connecting FG

Correspond to congruence of equilateral right triangle.

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So f is the midpoint of AG.

E is the midpoint of AD.

Therefore, in the triangle ADG, ef is the midpoint of both sides.

So EF//BC

3. From the second question, we can see EF//BC.

The extension line of EF bisects AB side and AC side respectively.

EF=(AC+BC-AB)/2

The last point can be extended from CF to CH to make FC=FH, which is proved by isosceles trapezoid and congruent triangles.