__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 ÒwrongÓ 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 AÕ, BÕ, and CÕ. By constructing segments connecting the midpoints, we have the medial triangle AÕBÕCÕ 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 ABÕCÕ = ½(½a)(½b) = 1/8ab

The
Area of triangle BÕCAÕ = ½(½a)(½b) = 1/8ab

The
Area of triangle CÕAÕB = ½(½a)(½b) = 1/8ab

The
Area of triangle AÕBÕCÕ = ½(½a)(½b) = 1/8ab

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

The triangle AÕBÕCÕ is constructed by the midpoints of the sides of triangle ABC. This medial triangle has base CÕAÕ which is parallel to segment AC. Because they are parallel, the perpendicular heights CÕG, NBÕ, and AÕP are parallel and congruent.

If we construct a circle with radius mNBÕ, and center at vertex B, we will see that the circle will intersect O, with radius BO. Thus mNBÕ = mBO which is the height of triangle CÕBAÕ. 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 letÕs construct a segment in this right triangle that
forms two triangles of equal area.

LetÕs construct the midpoints of each of the sides of triangle ABC. The midpoint of the hypotenuse CÕ 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 CCÕ, ACÕ, and BCÕ are congruent. Also, a segment formed by BÕCÕ is parallel to CB and a segment formed by AÕCÕ is parallel to AC. The segment CCÕ forms two other triangles: triangle CCÕB and triangle ACÕC. What are the areas of these triangles and their relationship to the main triangle ABC?

Area
of triangle CCÕB = ½(a)(½b) = ¼ab

Area
of triangle ACCÕ = ½(b)(½a) = ¼ab

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

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

Now letÕs 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 b_{1} = 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 AÕ,BÕ,CÕ, which are midpoints of the sides BC, CA, and AB respectfully. LetÕs prove that the circumcircles of triangles ABÕCÕ, BCÕAÕ, and CAÕBÕ have a common point.

LetÕs 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 ABÕCÕ, BCÕAÕ, and CAÕBÕ.

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.

LetÕs take a look at triangles that are not right.

With similar construction, we will look at triangle RST with midpoints RÕ, SÕ, and TÕ 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.