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Step-by-step explanation:

1. B, Concurrency of perpendicular

2. B, F is the infacility. D, F is the allude of concurrency of the angle. F, F is equidistant from the 3 sides of the.

3. B, right

4. B, (1.5, 1)

5. D, (3.5, 3)

6. A, isosceles

Part B: D, 122

7. C, 80

8. Part A: A, x=2

Part B: D, 7 units

As per the problem, we have actually been asked to find the name of the theorem which defines why the circum-facility is equiremote from the vertices of a triangle

A.Vertical Angles Theorem:

The vertical angle theorem is about the angles developed when two right lines reduced each other. Hence this is not the correct alternative

B.Conmoney of Perpendicular Bisectors Theorem

Perpendicular bisectors passing with the exact same point dubbed the circum-facility of the triangle. The suggest of intersection is always equidistant from the endpoints.

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Hence theorem "B.Concurrency of Perpendicular Bisectors Theorem" describes why the circumcenter is equifar-off from the vertices of a triangle.

C. Conmoney of Angle Bisector Theorem:

This gives you the incenter of the triangle.

D. Alternating Interior Angles Theorem

This theorem claims that if two parallel lines are intersected by a line then the different inner angles are congruent.

The proof is explained in step-by-step explaination.

Step-by-step explanation:

Circumcenter is the allude at which the perpendicular bisector of sides of triangle intersects inside the circle.

This points lies inside the triangle and circle and the vertices of triangles lies on the circle. As an outcome the distance from circumfacility and vertices is referred to as to be radius of the circle which is constantly equiremote from the center.

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Hence, circumcenter is equiremote from the vertices of a triangle.

The answer is B. Remember the definition of the circumcenter of a triangle: the intersection of perpendicular bisectors. If we attract a perpendicular line to each of the 3 sides, these three lines intersect at a common suggest, and the distance of this suggest to all 3 vertices are equal, which deserve to be proved using comparable triangles. If we draw a circle via the circumfacility of the triangle as the facility, and also with the distance from circumcenter to vertices as the radius, the triangle will certainly become inscribed in the circle.

The circumfacility is the center of the circle which goes with the three vertices of the triangle. Recall that all radii of a circle are congruent, i.e. equal to one another. So this is why the circumfacility is equifar-off from the vertices of the triangle.
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