The number of circles passing through the vertices of a triangle

Question

I have read a book written by C.V Durell on Geometry. In this book I have found a lemma which states that there is one and only one circle that passes through three vertices of a triangle.

I thought of proving by contradiction, but I just don't know how to go about.

Answer 1

1. Relax the definition of parallel so that a line may be said to be parallel to itself.

2. Prove that if lines $p_1$ and $p_2$ are parallel, and line $q_1$ is perpendicular to $p_1$ while line $q_2$ is perpendicular to $p_2,$ then $q_1$ is parallel to $q_2.$

3. Equivalent to 2.:If $q_1$ is not parallel to $q_2,$ and $p_1$ is perpendicular to $q_1,$ and $p_2$ is perpendicular to $q_2,$ then $p_1$ is not parallel to $p_2.$

4. Existence: Apply 3. to triangle ABC: The line thru $A,B$ is not parallel to the line thru $B,C$, so the right bisector $R_{A B}$ of $AB$ and the right bisector $R_{B C}$ of $BC$ are not parallel, so $R_{A B}$ and $R_{B C}$ intersect at some point $D.$ Since any point on $R_{A B}$ is equidistant from $A ,B,$ we have $AD=BD.$ Since any point on $R_{B C}$ is equidistant from $B, C$ we have $B D=C D.$ Hence $A D=B D= C D$ so $D$ is the center of a circle passing thru $A,B,C.$

5. Uniqueness: Any point $E$ which is equidistant from $A,B$ lies on $R_{A,B}$ because either (a) $E$ is the mid-point of $A B$, or (b) the altitude-line from $E$ to $AB$ in the isosceles triangle $AEB$ is $R_{A B}.$ Similarly, any point equidistant from $B,C$ lies on $R_{B C}.$ So if $E$ is equidistant from $A,B,C$ it must lie in the intersection of $R_{A B}$ with $R_{B C}.$ Since $R_{AB}$ and $R_{BC}$ are not parallel, they are not equal, so their only point of intersection is $D.$ So $E=D.$

Answer 2

The vertices A,B,C of the triangle form either 0, 1, or more triangles.

Take the perpendicular bisector of AB. Take the perpendicular bisector of BC. As AB is not parallel to BC. The perpendicular bisectors intersect at a point O. By definition of perpendicular bisector OA = OB and OB = OC. Thus A,B,C are on a circle centered at O with radius = OA =OB= OC.

Suppose there was another circle with points A,B,C and a center of O' and a radius of O'A= O'B = O'C.

As O'A = O'B, O' must be on the perpendicular bisector of AB and as O'B = O'C, O' must be on the perpendicular bisector of BC.

But two lines only intersect at one point. So if the perpendicular bisectors intersect at O AND at O', then O = O' and the radius O'A = OA, etc. and this is the only possible circle.

Thus for any triangle there is one and only one circle circumscribing the triangle.