Sum of Degrees in a Triangle
Date: 03/03/99 at 16:29:36 From: Anonymous Subject: A Triangle Proof Can you help me with the following homework problem? Prove that the sum of degrees in a triangle is 180 using a minimum of four different ways.
Date: 03/04/99 at 03:09:33 From: Doctor Floor Subject: Re: A Triangle Proof 1. Let ABC be a triangle, and draw a line through C parallel to AB (I added D and E for clarity): --D-----C--------E---- / \ / \ A--------B Because they are alternate angles, we see that angle(BCE) = angle(CBA) and angle(ACD) = angle(CAB). Since angle(BCE) + angle(BCA) + angle(ACD) = 180 degrees (they form a straight angle), the same goes for the angles of ABC. 2. Let ABC be a triangle. Construct a parallelogram on ABC in the following way: C--------D-----E / \ / / \ / A--------B Again, I added point E for clarity. Note that angle(BDE) = angle(ACD) = angle(ACB) + angle(ACD) = angle(ACB) + angle(ABC) and that angle(BDC) = angle(CAB). Since angle(BDE) + angle(BDC) = 180 degrees (straight angle), the same goes for the sum of the angles in triangle ABC. 3. Let ABC be a triangle. Let A' be the midpoint of BC, B' of AC and C' of AB. C / \ B'----A' / \ / \ A-----C'----B In this way we form four congruent triangles A'B'C', AB'C', A'BC' and A'B'C, of which the sum of the angles is equal to the sum of the angles of ABC. If we leave out the angles of triangle ABC, three straight angles are left for the sum of three of the triangles. So each triangle must have a sum equal to one straight angle = 180 degrees. 4. Let ABC be a triangle, and consider the following figure: \ \ * C / \ / \ * A--------B------ / * / Note that the three angles marked with * add up to one complete turn, i.e. 360 degrees. Note also that each of the angles marked with * makes a straight angle when added to one of the angles of ABC. So the three angles marked with * added to the angles of ABC add up to 3*180 = 540 degrees. That leaves 540 - 360 = 180 degrees for the angles of ABC. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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