Respuesta :

calculista

we know that

The circumcenter is the center of the the circle that passes through all three of the triangle's vertices.  Is the point of intersection of perpendicular bisectors of three sides of the triangle

so

[tex]ZL=ZM=ZN[/tex] -----> is equal to the radius of the circle  that passes through all three of the triangle's vertices

[tex]AN=AL[/tex]

[tex]BL=BM[/tex]

[tex]CN=CM[/tex]

In this problem

m∠ZBM=[tex]90\°[/tex]

m∠ZCM=[tex]90\°[/tex]

therefore

m∠ZBM=m∠ZCM -----> is true-

the answer is the option

m∠ZBM=m∠ZCM

Answer:

The correct option is 4.

Step-by-step explanation:

Given information: Point Z is the circumcenter of ΔLMN.

In a triangle the intersection point of perpendicular bisectors is called circumcenter.

Since Z is the circumcenter, therefore AZ, BZ, CZ are perpendicular bisectors.

[tex]AN=AL[/tex]

[tex]BM=BL[/tex]

[tex]CM=CN[/tex]

The distance between circumcenter and the vertices are equal.

[tex]LZ=MZ=NZ[/tex]

From the given figure it is noticed that

[tex]\angle LAZ=\angle NAZ=\angle NCZ=\angle ZCM=\angle ZBM=\angle LBZ=90^{\circ}[/tex]

Since we have

[tex]\angle ZCM=\angle ZBM[/tex]

Therefore option 4 is correct.