Proving equilaterial triangle using centroid and circumcenter

Printable View

December 10th 2009, 06:49 AM
ReneePatt

Proving equilaterial triangle using centroid and circumcenter

I am asked to: Prove that a triangle is equilaterial if and only if its centroid and circumcenter coincide.

I know that:

(a) centroid of a triangle is the point where the three medians (a segment joining a vertex of a triangle to the midpoint of the opposite side) of the triangle intersect.

(b) circumcenter of a triangle is the point where the three pependicular bisectors (a line segment that is both perpendicular to a side of a triangle and passes through its midpoint) of a triangle intersect.

But I don't know how to do the proof or what definitions/theorems are used for the proof.
December 10th 2009, 11:23 AM
Grandad

Hello ReneePatt

Quote:

Originally Posted by ReneePatt

I am asked to: Prove that a triangle is equilaterial if and only if its centroid and circumcenter coincide.

I know that:

(a) centroid of a triangle is the point where the three medians (a segment joining a vertex of a triangle to the midpoint of the opposite side) of the triangle intersect.

(b) circumcenter of a triangle is the point where the three pependicular bisectors (a line segment that is both perpendicular to a side of a triangle and passes through its midpoint) of a triangle intersect.

But I don't know how to do the proof or what definitions/theorems are used for the proof.

You have correctly identified the characteristics of the centroid and the circumcentre. You need to prove this in both directions.

First if the centroid and circumcentre coincide, prove that the triangle is equilateral. So:
Given: $\triangle ABC$ , and $L, M, N$ are the mid-points of $BC, CA, AB$ and $AL \perp BC, BM \perp CA, CN\perp AB$

To prove: $BC = CA=AB$

Proof:

In $\triangle\text{'s } ABL, ACL$ :
$BL = LC$ (given)

$\angle ALB = \angle ALC (=90^o$ , given $)$

$AL$ is common

$\therefore \triangle\text{'s } ABL, ACL$ are congruent. SSS.

$\therefore AB = AC$

Similarly in $\triangle\text{'s } BCM, BAM, BC = BA$

$\therefore BC = CA=AB$

Second we prove that if a triangle is equilateral, then the centroid and circumcentre coincide. Do this using congruent triangles to prove that each median makes an angle of $90^o$ with the opposite side. Can you complete this now?

Grandad
December 10th 2009, 02:09 PM
ReneePatt

Quote:

Originally Posted by Grandad

First if the centroid and circumcentre coincide, prove that the triangle is equilateral. So:

Given: $\triangle ABC$ , and $L, M, N$ are the mid-points of $BC, CA, AB$ and $AL \perp BC, BM \perp CA, CN\perp AB$

To prove: $BC = CA=AB$

Proof:

In $\triangle\text{'s } ABL, ACL$ :

$BL = LC$ (given)

$\angle ALB = \angle ALC (=90^o$ , given $)$

$AL$ is common

$\therefore \triangle\text{'s } ABL, ACL$ are congruent. SSS.

$\therefore AB = AC$

Similarly in $\triangle\text{'s } BCM, BAM, BC = BA$

$\therefore BC = CA=AB$

Second we prove that if a triangle is equilateral, then the centroid and circumcentre coincide. Do this using congruent triangles to prove that each median makes an angle of $90^o$ with the opposite side. Can you complete this now?

Grandad

Hi Grandad,

How do you have the givens? Are you assuming the givens because centroid and circumcenter are supposed to coincide?

Quote:

Originally Posted by Grandad

$BL = LC$ (given)

$\angle ALB = \angle ALC (=90^o$ , given $)$

Also, how are you getting that
$BL = LC$ (given)

$\angle ALB = \angle ALC (=90^o$ , given $)$

are given?
December 10th 2009, 03:59 PM
ReneePatt

I think I've got it

Grandad, I think I've got it - Thanks (Clapping)