### Video Transcript

Can we use point π in the given shape to draw a circle that passes through the vertices of triangle π΄π΅πΆ?

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. 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. Therefore, we must consider if the three line segments of line segment π΄π, π΅π, and πΆπ are the same length. If they were, then they would all be radii of the circle with center π. But we can see on the diagram that these are not the same length. And if this was a question on paper, we could measure the length with a ruler and establish that they are not the same. 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.

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. We can notice that the lines through each vertex in the triangle has bisected the angle at the vertex. 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. If we did want to find the center of a circle passing through these three points, we would use a perpendicular bisector. Letβs see how that would work in this triangle π΄π΅πΆ.

Letβs take a copy of triangle π΄π΅πΆ and remove the angle bisectors. The method here will be to take the perpendicular bisectors of two of the line segments. So letβs start with line segment π΄π΅. If we want to accurately construct a perpendicular bisector, we use a compass. 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. This line represents all the points which are equidistant from point π΄ and π΅.

We can follow the same process to find the perpendicular bisector of the line segment π΅πΆ. This perpendicular bisector will give us all the points that are equidistant from points π΅ and πΆ. And we notice that there is a point of intersection between these two perpendicular bisectors. This point is equidistant from π΄, π΅, and πΆ. 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.

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. We do not use the angle bisectors. 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.