section6

ANGLES IN A TRIANGLE, RIGHT TRIANGLES AND EQUALITY OF RIGHT TRIANGLES BY HYPOTENUSE-LEG, EQUALITY OF TWO TRIANGLES BY SIDE-SIDE-SIDE.

In this section we are going to finish our discussion of triangles and their properties. One of the remarkable results regarding angles in any triangle is formulated in the following theorem.
Theorem 6.1 Sum of degree measures of all angles in a triangle is equal to 180 degrees.
Proof Let ABC be an arbitrary triangle . We need to prove that

Let's construct a line p such that it will pass through the point C and will be parallel to AB. Let D be a point on p located right to the point C. By construction angles DCB and CBA are equal as alternate interior angles for parallel lines p and AB and their transversal CB.

At the same time angles DCA and CAB are one-sided interior angles for parallel lines p and AB and their transversal AC.

Therefore

Finally,

So, we get the following system of equations:

Now we are going to consider one more type of triangles, so called right triangles. A triangle is called a right triangle if one of its angles is equal to 90 degrees. From the theorem 6.1 it follows that the other two angles must sum up to 90 degrees, therefore each of the other two angles should be acute.

In a right triangle, the side opposite the right angle is called the hypotenuse. The other two sides are called legs.

Theorem 6.2 (the equality of two right triangles by hypotenuse-leg, HL) If a hypotenuse and a leg of one right triangle is equal to a hypotenuse and a leg of another triangle then these two triangles are equal.

Proof Suppose that we have two right triangles ABC and DEF with AB = DE, AC = DF and

We need to show that triangles ABC and DEF are equal. Let's construct a triangle CBH equal to the triangle DEF and a triangle EFI equal to the triangle ABC.

Triangles ABH and DEI are equal by SSS (see Theorem 6.3 below): AB = DE, AH = AC+CH=FI+DF = DI, BH = DE = AB = EI. Therefore

So, the triangles ABC and DEF are equal by SAS: AB = DE, AC = DF,

The next step that we are going to make will be to prove one more theorem regarding the equality of two triangles.

Theorem 6.3 (the equality of two triangles by side-side-side, SSS) If three sides of the triangle ABC are equal to three sides of the triangle DEF then triangles ABC and DEF are equal.
Proof Suppose that AB = DE, BC = EF, AC = DF.

While proving this theorem we are going to introduce one more proving technique: a proof by contradiction. The main idea of this technique is to assume that the statement that we want to prove is false. After this assumption is made we are going to show that in a few logical steps we'll get a contradiction of our assumption and some result that is either a theorem that has been already proved or an axiom.

Suppose that triangles ABC and DEF are not equal. Then
Suppose that DGF is a triangle that is equal to the triangle ABC (such a triangle can always be constructed according to the axiom 3.) Let H be the middle point of the segment EG. Then the triangles EGF and EGD are isosceles triangles (DG = DE, EF = GF) with the same base side EG. Medians FH and DH are also altitudes by the theorem 5.3. Therefore two distinct lines DH and FH are both perpendicular to the line GE and they both pass through the same point H!

But this result contradicts the theorem 3.1 that says that given a point A and a line m that contains A we can always construct a line n such that it will be perpendicular to m and that will pass through A. Moreover, such line n will be unique. Therefore our assumption that triangles ABC and DEF are not equal was false, so these triangles are indeed equal.

EXERCISE SET #6

Problem 1. Find the degree measure of a vertex angle in an isosceles triangle if the base angle in this triangle is equal to 20 degrees.

Problem 2. Can lines AB and CD be parallel if

Why?