Respuesta :

To prove:

[tex]\Delta\text{CAM}\cong\Delta\text{DBM}[/tex]

Statement: 1

[tex]\angle ACM\cong\angle\text{BDM}=90^{\circ}[/tex]

Reason: Given

Statement:2

[tex]\angle CMA\cong\angle DMB[/tex]

Reason:

Angle 1 and angle 2 are congruent angles.

And also,

[tex]\begin{gathered} \angle CMA\cong\angle1(\text{Alternate angles)} \\ \angle DMB\cong\angle2(\text{Alternate angles)} \end{gathered}[/tex]

Statement: 3

[tex]AM=MB[/tex]

Reason: M is the midpoint of AB.

Statement: 4

[tex]\Delta\text{CAM}\cong\Delta\text{DBM}[/tex]

Reason: By AAS congruence.

Hence proved.

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