Finding a Point That is Equidistant from 3 Other Points in a Triangle

[bobron]

I have a math problem in my geometry homework that states the following:
Below is triangle △FPT. Do you believe there is any point equidistant from P, F, and T? If so, copy the triangle and draw the point E and state the distance from F, P, and T to E.

I know that I probably have to use a compass for it, but how would I?

 

[srmichael]

I have a math problem in my geometry homework that states the following:
Below is triangle △FPT. Do you believe there is any point equidistant from P, F, and T? If so, copy the triangle and draw the point E and state the distance from F, P, and T to E.
View attachment 3144
I know that I probably have to use a compass for it, but how would I?

I assume you are learning about orthocenters, incenters, centroids and circumcenters. Do you know some of the special qualities about some of these?

 

[pka]

I have a math problem in my geometry homework that states the following:
Below is triangle △FPT. Do you believe there is any point equidistant from P, F, and T? If so, copy the triangle and draw the point E and state the distance from F, P, and T to E.
View attachment 3144
I know that I probably have to use a compass for it, but how would I?

Draw the perpendicular bisectors of $\displaystyle \overline{PT}~\&~\overline{FT}$.
Their intersection point is what you seek.

 

[bobron]

Draw the perpendicular bisectors of $\displaystyle \overline{PT}~\&~\overline{FT}$.
Their intersection point is what you seek.

Thanks a lot! That solved my question.
But, could I ask why it's like that? I really want to know.

 

[pka]

But, could I ask why it's like that? I really want to know.

It is a fact that any three non-collinear points determine a circle.
Given two points, the perpendicular bisector of the line segment they determine is the locus of all points which are equally distance from the two points. Finding two bisectors that intersect gives the center of the circle.