Triangles—Right or Wrong

Many times in the secondary classroom, students develop their own theories of shapes and the properties thereof. Properties of triangles can easily be generalized with the uses of right triangles. In the following examples, I will demonstrate two different properties of triangles involving right and “wrong” triangles.

Given a right triangle ABC.

How would we create four triangles of equal area within this triangle?

We will first find the midpoints of each side of the triangle and labeled A’, B’, and C’. By constructing segments connecting the midpoints, we have the medial triangle A’B’C’ which is also a right triangle.

Each of these segments have the given a length mAC = b, mCB = a, mAB = c.

If the four triangles are congruent, then they all have equal areas and are 1/4^th the area triangle ABC.

The Area of triangle ABC = ½bh = ½ab

The Area of triangle AB’C’ = ½(½a)(½b) = 1/8ab

The Area of triangle B’CA’ = ½(½a)(½b) = 1/8ab

The Area of triangle C’A’B = ½(½a)(½b) = 1/8ab

The Area of triangle A’B’C’ = ½(½a)(½b) = 1/8ab

Now let us look at triangles that are not quite right.

The triangle A’B’C’ is constructed by the midpoints of the sides of triangle ABC. This medial triangle has base C’A’ which is parallel to segment AC. Because they are parallel, the perpendicular heights C’G, NB’, and A’P are parallel and congruent.

If we construct a circle with radius mNB’, and center at vertex B, we will see that the circle will intersect O, with radius BO. Thus mNB’ = mBO which is the height of triangle C’BA’. Therefore, the triangle ABC is cut into four congruent triangles.

As a challenge, express the area of each of the four congruent triangles in terms of a, b, and c.

Now let’s construct a segment in this right triangle that forms two triangles of equal area.

Let’s construct the midpoints of each of the sides of triangle ABC. The midpoint of the hypotenuse C’ is equidistant from all three of the vertices of triangle ABC because the center of a circumscribed right triangle is the midpoint of the hypotenuse. Therefore, the segments CC’, AC’, and BC’ are congruent. Also, a segment formed by B’C’ is parallel to CB and a segment formed by A’C’ is parallel to AC. The segment CC’ forms two other triangles: triangle CC’B and triangle AC’C. What are the areas of these triangles and their relationship to the main triangle ABC?

Area of triangle CC’B = ½(a)(½b) = ¼ab

Area of triangle ACC’ = ½(b)(½a) = ¼ab

¼ab + ¼ab = 1/2ab è Area of triangle ABC.

Thus, this segment CC’ divides the triangle ABC into two equal areas.

Now let’s look at triangles that are not quite right.

Suppose we are given a non-equilateral triangle WXZ and take the midpoint R on the hypotenuse, creating the segment XR.

Here we create two different triangles WXR and ZXR. In these new triangles, we need to determine the altitudes to determine the areas.

In order to have equal areas, the Area of WXR = Area of ZXR

1/2bh = 1/2bh

If we draw an altitude from X to the base WY, this height will be the same for both triangles. Our bases for our two triangles are composed of WZ. Therefore, if we have equal heights h then we must have equal bases where b₁ = b_2. Therefore, if our heights and bases are the same then we have two triangles with equal areas.

Triangles and Circles

Given a triangle ABC and points A’,B’,C’, which are midpoints of the sides BC, CA, and AB respectfully. Let’s prove that the circumcircles of triangles AB’C’, BC’A’, and CA’B’ have a common point.

Let’s explore the right triangle ABC

When constructing the circumcircles of this triangle, we must find the intersection of each of the perpendicular bisectors of triangles AB’C’, BC’A’, and CA’B’.

In the right triangle ABC, all four triangles created, including the medial triangle, are right as well. The circumscribed circles of these triangles all have midpoints of the hypotenuse as the centers. In right triangles, the intersection of these constructed triangles is the midpoint of the hypotenuse of triangle ABC.

Let’s take a look at triangles that are not right.

With similar construction, we will look at triangle RST with midpoints R’, S’, and T’ of ST, RT, and SR respectively.

When the medial triangle is formed, we have 4 congruent triangles, including the medial triangle. The three circumscribed triangles all have equal radii, indicating they have one point within the triangle equidistant from all three vertices. The triangle formed by C1, C2, and C3 is congruent to the 3 circumscribed. Also, this triangle has one center, in which it is the center of a circumscribed circle. The center of that circle is the only equidistant point inside the triangle, which is the only point where the circles C1, C2, and C3 will intersect.