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Can we use point π in the given shape to draw a circle that passes through the vertices of triangle π΄π΅πΆ?
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Letβs begin by remembering that a circle is a set of points in a plane that are a constant distance from a point in the center.
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And that means that if point π is the center of the circle, then π΄, π΅, and πΆ will all be a constant or the same distance from the center π.
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Therefore, we must consider if the three line segments of line segment π΄π, π΅π, and πΆπ are the same length.
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If they were, then they would all be radii of the circle with center π.
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But we can see on the diagram that these are not the same length.
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And if this was a question on paper, we could measure the length with a ruler and establish that they are not the same.
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And so the short answer to this question is no, we couldnβt use point π to draw a circle passing through the vertices of triangle π΄π΅πΆ.
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But letβs have a look at what has been done in the diagram and how we could find the center of a circle which passes through π΄, π΅, and πΆ.
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We can notice that the lines through each vertex in the triangle has bisected the angle at the vertex.
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This type of trick is often used in an exam question because we do need some sort of bisector to find the center of a circle of these three vertices, but itβs not an angle bisector.
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If we did want to find the center of a circle passing through these three points, we would use a perpendicular bisector.
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Letβs see how that would work in this triangle π΄π΅πΆ.
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Letβs take a copy of triangle π΄π΅πΆ and remove the angle bisectors.
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The method here will be to take the perpendicular bisectors of two of the line segments.
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So letβs start with line segment π΄π΅.
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If we want to accurately construct a perpendicular bisector, we use a compass.
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Placing our compass point in points π΄ and π΅ in turn and tracing some arcs on either side of the line segment, we can then join a line through the two points of intersection of the arcs.
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This line represents all the points which are equidistant from point π΄ and π΅.
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We can follow the same process to find the perpendicular bisector of the line segment π΅πΆ.
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This perpendicular bisector will give us all the points that are equidistant from points π΅ and πΆ.
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And we notice that there is a point of intersection between these two perpendicular bisectors.
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This point is equidistant from π΄, π΅, and πΆ.
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The three line segments from the point of intersection to each vertex on the triangle could be the radii of a circle, a circle which might look something like this.
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And so weβve demonstrated that to find the center of a circle which passes through three points, we find the perpendicular bisectors of two line segments.
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We do not use the angle bisectors.
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Using angle bisectors as was done here to find the point π means that point π cannot be used to draw a circle which passes through the vertices of triangle π΄π΅πΆ.