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

E. C. Titchmarsh (1899-1963)

First it is important to understand the concept of

To proceed, click on each

Proof:

Step 1:

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

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

Construct any point K on the perpendicular bisector p.

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.

Construct the midpoint of segment AC. Call it M'.

Construct the perpendicular bisector of AC. Label as

Since AC and BC are not parallel, the perpendicular bisectors, **p**
and **q**, must also intersect.

Let point K be the point of intersection.

Constuct point K as the point of intersection of

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.

This proof can extend to other concepts such as the

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.