#### Many of the proofs in mathematics are very long and intricate. Others, though not long, are very ingeniously constructed.

E. C. Titchmarsh (1899-1963)
We want to show that the three perpendicular biectors of the sides of a triangle are concurrent.

First it is important to understand the concept of concurrency.

Concurrent lines (segments or rays) are lines which lie in the same plane and intersect in a single point. The point of intersection is the point of concurrency. For example, point A is the point of concurrency.

The following is an interactive proof using GSP.

To proceed, click on each Step of the proof. If you want to retrace your steps you may undo your previous step by clicking on the Undo Step.

Claim:

### Step 2

Construct the midpoint of segment AC. Label the midpoint of AC as M.

### Step 3

Construct the perpendicular bisector of segment AC. Call the perpendicular bisector p.

### Step 4

Construct any point K on the perpendicular bisector p.

### Step 5

Construct segment AK and segment KC.

AM=MC by definition of midpoint.
KM=KM by the reflexive property.
Angle AMK and angle CMK are cut by a perpendicular line, p. Hence, the angles are right angles and are both equal to 90 degrees. So, m<AMK = m<CMK and angle AMK is congruent to angle CMK.
Now, triangle AMK is congruent to triangle CMK by SAS.
By CPCTC, AK=KC.

### Step 6

Construct the midpoint of segment AC. Call it M'.
Construct the perpendicular bisector of AC. Label as q.

Since AC and BC are not parallel, the perpendicular bisectors, p and q, must also intersect.
Let point K be the point of intersection.

### Step 7

Constuct point K as the point of intersection of p and q.

### Step 8

Construct segment BK.

KC=BK using similar reasoning in Step 5. We know AK=KC. So by substitution AK=KC=BK.
Furthermore, K is taken to be the intersection of the perpendicular bisectors p and q.
Since K is on the perpendicular bisectors p and q, it must also lie on the perpendicular bisector of AB.
Therefore, K is on the perpendicular bisector of AB.

Since K lies on the perpendicular bisectors of AC, BC, and AB, K is the point of concurrency.
Hence, we have proved that the three perpendicular bisectors of the sides of a triangle are concurrent.

The following shows the construction of the perpendicular bisector of AB, labeled as r.
The picture shows K to be the point of concurrency.

## Extension

This proof can extend to other concepts such as the circumcenter and circumcircle.

K, the point of concurrency, is also the circumcenter. The circumcenter is the center of the circumcircle. The circumcircle is the circumscribed circle of the triangle.

The following shows the cirumcircle of triangle ABC with circumcenter as point K.