Review for Chapter 10 Test (Day 143)

Last night I tutored my geometry student. Section 7-5, as it turned out, was the final section of my student's edition of the Glencoe text. So today he is reviewing for the Chapter 7 Test.

Also, as it turned out, his teacher didn't really cover the Angle Bisector Theorem in Section 7-5. The focus instead was on how altitudes, angle bisectors, medians, and perimeters of similar triangles are in the same proportion as the sides. In the U of Chicago text, this fact for perimeters is part of the Fundamental Theorem of Similarity. Recall that one of our links from the beginning of the year, Jen Silverman, had a CPCTC statement that included the altitudes of the triangles, and Glencoe extends this idea to from congruence to similarity. It's easy to prove that if two triangles are similar, then the altitudes have the same scale factor, at least in the case where the angles are acute (so that the altitudes are inside the triangle). A two-column proof:

Given: Triangle RST ~ EFG, SA and FB are altitudes
Prove: SA/FB = ST/FG

Proof:
Statements                                Reasons
1. RST ~ EFG, SA, FB altitudes  1. Given
2. Angle T = Angle G                 2. Similar Figures Theorem (as applied to angles)
3. Angles SAT, FBG right angles 3. Definition of altitude
4. Angle SAT = FBG                  4. All right angles are congruent.
5. Triangle SAT ~ FBG              5. AA Similarity Theorem
6. SA/FB = ST/FG                      6. Similar Figures Theorem (as applied to sides)

Notice that if the two triangles are obtuse, the altitudes are outside the triangle. So we need an extra step the (acute) exterior angles, the supplements of the obtuse angles, to be congruent, as these are the second angles of the right triangles. The proofs for angle bisectors and medians work the same way -- each segment divides the large pair of similar triangles into two smaller pairs of triangles that also turn out to be similar.

Now the question on which my student struggled the most involved the two similar triangles RST and EFG, with altitudes SA and FB, with lengths given as follows: SA = 2, ST = x, FB = 7 - x, FG = 5, and the student was asked to find FB. The student set up the proportion, 2/(7 - x) = x/5, but failed to notice that cross-multiplying led to a quadratic equation with two solutions. The two solutions of this equation happen to be x = 2 and x = 5. But the solution x = 2 leads to nonsense -- if ST = x = 2 = SA, then SAT is a right triangle whose hypotenuse is congruent to one of its legs. Therefore only x = 5 is a viable solution and FB = 2. Notice that this makes the similar RST and EFG be congruent triangles.

Like my student, we are reviewing for a test today, too.