# Kenneth E. Montgomery

Problem

Construct a triangle and its medians. Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. (E.g., are they congruent? similar? Have same area? same perimeter? ratio of areas? ratio or perimeters?) Prove whatever you find.

Overview

Relationships between the segments and angles formed in the construction of a triangle and its medial triangle are investigated. This problem was explored using the Key Curriculum Press, Geometer’s Sketchpad ® software. Questions regarding the ratio of areas and perimeters as well as similarity and congruency of angles, sides and triangles are explored using the software. Observations are made and hypotheses tested, within the exploration, and a proof of each result is offered below.

Investigation

Upon construction of  , the midpoints (DE and F) as are the connecting segments,  and  , (Figure 6.1). It should be noted that individual segments are also constructed between points A and D, et cetera, so that measurements of each individual segment within the triangle may be found.

Figure 1:    Scalene with medial

Open this GSP file to explore the properties of the triangle in Figure 1. The length of a segment may be measured by clicking on the desired segment and selecting Length from the Measure menu on the toolbar. The observed measurements are presented in Table 6.1.

Table 6.1

From these measurements it is clear that, for this specific triangle at least, congruence relationships exist. Further, we hypothesize congruence relationships for angles. To investigate, it is necessary only to select three points, which together represent an angle (e.g. DB and F for  ). Once these points are selected, we select Angle from the Measure menu. The hypotheses should be verified, again for this triangle in particular, by observation of the measurements in Table 6.2.

Table 6.2

An understanding of congruent triangles should be developed from this exploration. We may at this time conjecture that  . This investigation helps to distill an argument for why the original triangle ( ) is not congruent to the medial triangle ( ). Further investigation of segment length in  produces the data in Table 6.3. A comparison of these values to those in Table 6.1 provide a clear argument as to why the two triangles are not congruent: corresponding side lengths are not congruent.

Table 6.3

What we notice, however is that the ratio of side lengths is 2:1 and this is also verified in Sketchpad® by selecting two corresponding segments, such as  and  and selecting Ratio from the Measure menu. The ratio is verified for all three pairs of angles. We conjecture that the triangles are similar and by any of the three theorems for similarity (SSS, SAS, ASA) we can empirically verify that they are.

The question that arises is whether any of our hypotheses are true in general, or just for this triangle. Again, Sketchpad® may be used to investigate, by either constructing different triangles (and repeating our previous steps) or by dynamically changing the properties of our original triangle, while measuring each of the quantities mentioned. A GSP file is provided as an illustration. However, while we have verified the above-mentioned properties for more triangles than just our original, we have not yet proven this in general, nor have we addressed the issues of perimeter or area.

Note to self: come back and fix fonts.

Proof:

Given  , with constructed midpoints and segments, we have the medial triangle,  (Figure 1). Since D is the midpoint of AB, F is the midpoint of BC and E is the midpoint of AC. It follows that AD=DB, BF=FC, and BE=EC.

By the SSS Theorem (Side, Side, Side), we have  . By the CPCTC Theorem (Corresponding Parts of Congruent Theorems are Congruent) we also have congruent angles:  and  as well as

The Definition of Similar Polygons states that two polygons are similar if and only if their corresponding angles are congruent and the measures of their corresponding sides are proportional. Therefore, since

Proof:

Part 1:

Given  , with constructed midpoints and segments, we have the medial triangle,  (Figure 1). Since  and  , it follows that  and

Since  and , we have and  . By the reflexive property, we have and  by SAS similarity. Ò

Part 2:

Since  , we have  and without loss of generality, we also have and . Thus and , with (reflexive property). By SSS congruence we thus have .Ò

Part 3:

Ò

Part 4:

Because the larger triangle ( ) is composed of the smaller triangles, we have:

Since,

the areas of each triangle are equal:

By the transitive property, we have:

therefore,

Ò

Return to Homepage