Question 103
The perpendicular bisector of a segment $X Y$ is the line perpendicular to the segment passing through its midpoint. The main property of the perpendicular bisector is that a point lies on it if and only if the distance from this point to $X$ equals the distance from this point to $Y$. A statement formed with "if and only if" is equivalent to two implications. In the present case, to verify the statement, we must on the one hand show that if the distance to $X$ equals the distance to $Y$, then the point lies on the perpendicular bisector. On the other hand, we must show that if a point lies on the perpendicular bisector, then its distances to the points $X$ and $Y$ are equal. To prove the first part, consider a point $P$ such that $P X=P Y$ and let $M$ be the midpoint of $X Y$.
Triangles $P M X$ and $P M Y$ are congruent, since their three corresponding sides have the same lengths. Therefore $\angle P M X=\angle P M Y$. Since $X, M$ and $Y$ are collinear, we have $\angle P M X+\angle P M Y=180^{\circ}$. Consequently $\angle P M X=90^{\circ}$ and $P$ lies on the line perpendicular to the segment through its midpoint.
a) Prove the second part of the proposition, that is, prove that if $P$ is a point on the perpendicular bisector of $X Y$, then $P X=P Y$.
b) Prove that every triangle $A B C$ has a point $O$ in the same plane such that $O A=O B=O C$. This point is called the circumcenter of the triangle, since a circle with center $O$ and radius $A O$ passes through the three vertices of triangle $A B C$.
c) Consider a quadrilateral $A B C D$. Suppose that the perpendicular bisectors of $A B, A C$ and $A D$ are concurrent, that is, the three pass through a common point. Prove that there is a circle passing through the 4 vertices of the quadrilateral $A B C D$.
Triangles $P M X$ and $P M Y$ are congruent, since their three corresponding sides have the same lengths. Therefore $\angle P M X=\angle P M Y$. Since $X, M$ and $Y$ are collinear, we have $\angle P M X+\angle P M Y=180^{\circ}$. Consequently $\angle P M X=90^{\circ}$ and $P$ lies on the line perpendicular to the segment through its midpoint.
a) Prove the second part of the proposition, that is, prove that if $P$ is a point on the perpendicular bisector of $X Y$, then $P X=P Y$.
b) Prove that every triangle $A B C$ has a point $O$ in the same plane such that $O A=O B=O C$. This point is called the circumcenter of the triangle, since a circle with center $O$ and radius $A O$ passes through the three vertices of triangle $A B C$.
c) Consider a quadrilateral $A B C D$. Suppose that the perpendicular bisectors of $A B, A C$ and $A D$ are concurrent, that is, the three pass through a common point. Prove that there is a circle passing through the 4 vertices of the quadrilateral $A B C D$.