Centers of a Triangle.

By

Ronnachai Panapoi

In this assignment, we had an opportunity to use GSP in exploring some basic triangle geometry. After examining the activities, I decided to choose some fact and write it in the form of proof.

“Prove that the three perpendicular bisectors of the sides of a triangle are concurrent”

We will prove that these three perpendicular bisectors meet at the point P.

Proof:

Click HERE to investigate the fact about the perpendicular bisector before doing the proof.

Let PD and PE be the perpendicular bisectors of segments AB and BC, respectively.

In this case, we have otherwise BA, BC would be collinear.

Draw a segment from a midpoint F of AC meet DP at the point P.

We will show that PF is perpendicular to AC.

Construct segments AP, BP, and CP.

We have AD = DB, , PD = PD.

So, (SAS congruence)

Therefore, AP = BP

Similarly, we have .

This is because BE = BE, , PE = PE.

Hence, BP = CP.

So, AP = BP = CP.

We now have AP = CP, AF = FC, and PF = PF.

Thus, (SSS congruence).

So, since .

This suggests that PF is perpendicular to AC.

So, we completely prove that the three perpendicular bisectors of the sides of a triangle are

concurrent.

RETURN