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.
Written mathematical problems are usually mathematical problems that need to be calculated on a draft with the help of a pen. You need to write down the calc