Hello! We can prove this with some statements, look:
• Notice that CD is parallel to AB;
,
• angle 1 ≅ angle 2
,
• M is the midpoint of AB;
Statement 1:
[tex]\hat{\text{AMC}}\cong\hat{\text{BMD}}[/tex]
Reasoning 1:
As M is the midpoint of AB and we have two similar lines starting in M, we can divide the angle M into two equal angles m1 and m2.
definition of midpoint
Statement 2:
[tex]\hat{C}=\hat{D}[/tex]
Reasoning 2:
As we already know that angle 1 ≅ angle 2, let's calculate the sum of the angles C and D, look:
angle C:
90º + angle 1
angle D:
90º + angle 2
so, 90º + angle 1 ≅ 90º + angle 2, it means that angle C ≅ angle D.
Statement 3:
[tex]\hat{MAC}\text{ = }\hat{\text{MBD}}[/tex]
Reasoning 3:
The sum of the internal angles of a triangle must be equal to 180º, right? So, knowing it we can say that angles A and B are equal, look:
m1 + A + C = m2 + B + D = 180º
remember, m1 ≅ m2 and C ≅ D, so A ≅ B too.
According to the explanation and image, we can prove that triangle CAM ≅ triangle DBM.
Table:
