Monday Math 150

Prove:
1) That all three medians of a triangle intersect at a single point (the centroid of the triangle), and that this point divides the medians into segments with a 2:1 length ratio.

and

2) That the six smaller triangles into which a triangle is divided by its medians have equal area.

1. Let us consider a triangle ∆ABC. Let points M, N and O be the midpoints of the line segments BC, AC and AB, respectively. Then, let P be the intersection of the medians AM and BN. Lastly, let S be the point where the line through M parallel to BN intersects CN.

CentroidFig1

By the corresponding angles postulate, ∠CMS≅∠CBN and ∠CSM≅∠CNB. Thus, by the AA postulate, ∆CMS~∆CBN, and so CS:CN=CM:CB. But since M is the midpoint of BC, CM is one half the length of BC. Thus, CS is one half the length of CN; and S is the midpoint of CN. Thus NS is also one half the length of CN, and since N is the midpoint of AC, and so AN=CN, we see that AN:NS=CN:NS=2:1.

By the corresponding angles postulate, ∠APN≅∠AMS and ∠ANP≅∠ASM, and so ∆APN~∆AMS as well. Therefore, AM:AP=AS:AN, and since PM=AM-AP and NS=AS-AN, PM:AP=(AM:AP)-1=(AS:AN)-1=NS:AN, and so we then see that AP:PM=AN:NS=2:1.

By similarly constructing a line segment from point N to MC parallel to AM, we can establish BP:PN=2:1 as well. And by similarly considering the medians AM and CO, we can analogously prove that their point of intersection must divide AM into segments with a 2:1 ratio, which is the point P. And thus, we see that AM, BN, and CO all intersect at point P, with AP:PM=BP:PN=CP:PO=2:1.

2. For any triangle, a median divides it into two smaller triangles with equal areas. To see that is is true, consider ∆ABC with median AM and altitude AE.

CentroidFig2

Since M is the midpoint of BC, BM=MC. Further, AE is an altitude of both ∆ABM and ∆ACM as well. Thus, with BM and MC as the bases of ∆ABM and ∆ACM respectively, we see the corresponding heights of these trianges are both the length of AE. Since they have equal length bases and equal heights, the triangles have equal area.

Now consider a triangle ∆ABC with medians AM, BN and CO, and centroid P, as in part (1). Then AM divides the area of ∆ABC in half: area(∆ABM)=area(∆ACM).
Similarly, PM is a median of triangle ∆PBC, and so area(∆PBM)=area(∆PCM).

CentroidFig3

Now, since area(∆ABP)=area(∆ABM)-area(∆PBM) and area(∆ACP)=area(∆ACM)-area(∆PCM), we see then that area(∆ABP)=area(∆ACP). Analogous arguments can show that area(∆ABP)=area(∆BCP) and area(∆BCP)=area(∆ACP), and thus area(∆BCP)=area(∆ACP)=area(∆ABP)=1/3*area(∆ABC). Since PM, PN and PO divide in half the areas of ∆BCP, ∆ACP and ∆ABP, respectively, we see area(∆AOP)=area(∆BOP)=area(∆BMP)=area(∆CMP)=area(∆CNP)
=area(∆ANP)=1/6*area(∆ABC).

Advertisements

Tags: , , , ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: