Yes in equilateral triangle.
In a plane, there are infinitely many lines that can serve as perpendicular bisectors of a given segment. The unique perpendicular bisector of a segment is a specific line that divides the segment into two equal parts at a right angle. However, any line parallel to this unique bisector, at any distance, can also be considered a perpendicular bisector if it intersects the segment at its midpoint. Thus, while the unique perpendicular bisector exists, an infinite number of lines can be drawn parallel to it.
A segment has exactly one bisector. This bisector is a line (or line segment) that divides the original segment into two equal parts and is perpendicular to it. No matter the length of the segment, the unique bisector will always pass through the midpoint of the segment.
A circle !!!!!!
It could be a right angle triangle
It could be but does not have to be... (Perpendicular to what?)
The description seems a bit confusing (to me) but it sounds like it could be a perpendicular bisector of a side of a triangle.
The median is a line from a vertex to the midpoint of the opposite line and an altitude is a line from a vertex to the opposite line which is perpendicular to the line. These are NOT the same thing in most triangles. The only time they could be the same is in an equilateral triangle.
If the compass angle is changed, the entire geometric shape being drawn is different. For example, if a triangle is being drawn, it could change from an obtuse triangle from a ninety degree triangle.
In a plane, there are infinitely many lines that can serve as perpendicular bisectors of a given segment. The unique perpendicular bisector of a segment is a specific line that divides the segment into two equal parts at a right angle. However, any line parallel to this unique bisector, at any distance, can also be considered a perpendicular bisector if it intersects the segment at its midpoint. Thus, while the unique perpendicular bisector exists, an infinite number of lines can be drawn parallel to it.
It could be, but without the diagram it is not possible to be certain.
You have points A, B, and C. Using a compass and straight edge, find a perpendicular bisector of AB (that is, a line that is perpendicular to AB and intersects AB at the midpoint of AB. Next, find a perpendicular bisector of BC. The two lines you found will meet at the center of the circle.
In the same way that you bisect an acute triangle. Alternatively, you could extend one of the rays of the obtuse angle so that you have an acute angle. Bisect that angle and then draw a perpendicular to the bisector of the acute angle through the vertex.
The median of an isosceles triangle from its apex is also the perpendicular bisector of the base. This line divides the triangle into two congruent right angled triangles whose hypotenuse is 3 feet and whose apical angle is 35/2 = 17.5 degrees. If the base of the original triangle was 2b cm then sin(17.5) = b/3 so that b = 3*sin(17.5) = 0.9cm so that the base was 2b = 1.8 feet Alternatively, you could use the sine rule on the triangle:
A segment has exactly one bisector. This bisector is a line (or line segment) that divides the original segment into two equal parts and is perpendicular to it. No matter the length of the segment, the unique bisector will always pass through the midpoint of the segment.
Not necessarily. It is a plane shape which could be horizontal.
A circle !!!!!!
It could be a right angle triangle