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Mathematical Triangle Proof (Grade One)
1)

Method 1: add a condition: ∠ABD=∠BAC.

Yes, ∠ 1=∠2, ∠ABD=∠BAC, AB=AB.

Therefore, triangle ABD is congruent with triangle BAC.

AC=BD

Method 2: add a condition: ∠ c = ∠ d.

I see, ∠ 1=∠2,

Get: ∠ABD=∠BAC

The following is the same as method 1.

Method 3: add a condition: AD=BC.

Yes, ∠ 1=∠2, AB=AB.

Therefore, triangle ABD is congruent with triangle BAC.

AC=BD

2)

Conclusion:

One: AC is perpendicular to BD.

Because AD=AB and CD=CB.

Then, AC is the symmetry axis of ABCD, and B and D are symmetrical about AC.

Lines b and d are perpendicular to the axis of symmetry AC.

Two: BE=DE

Because AD=AB, AC is perpendicular to BD.

AE divides BD equally, BE=DB.

Three: ∠DAE=∠BAE

Because AD=AB, AC is perpendicular to BD.

So AE is the bisector of the bad vertex angle.

∠DAE=∠BAE

3)

Method 1: Add condition: ∠CAB=∠DAB.

Because AC=AD, AB=AB, ∠CAB=∠DAB.

Therefore, triangle ABC is congruent with triangle ABD.

Method 2: Conditional: BC=BD

Because AC=AD, AB=AB, BC=BD.

Therefore, triangle ABC is congruent with triangle ABD.

Method 3: add conditions, CE=DE.

Because, AC=AD, AE=AE, CE=DE.

Therefore, the triangle ACE is congruent with the triangle ADE.