Homework #3 - Solutions


Problem 1: Using the SAS criteria for similarity (see the first page), prove the `midsegment theorem': Let $\Delta ABC$ be a triangle and suppose that $D$ is the midpoint of $\overline{AB}$ and $E$ is the midpoint of $\overline{BC}.$ Prove that MATH is parallel to MATH and $DE=\frac{1}{2}AC.$

Solution: First note that since $D$ is the midpoint of $\overline{AB}$ and $E$ is the midpoint of $\overline{BC},$ we have that
MATH
Since $\angle B$ is congruent to itself, the SAS criteria for similarity yields that MATH. Now $\angle BAC$ is congruent to $\angle BDE$ and hence (as MATH is transversal to MATH and MATH and forms a pair of corresponding congruent angles) MATH is parallel to MATH The similarity also yields that $DE/AC=1/2$ and hence $DE=\frac{1}{2}AC.$


$\medskip $Problem 2 :Prove the Law of Sines: If $\Delta ABC$ is any triangle, then
MATH

In order to prove this, let $D$ be the foot of the perpendicular dropped from $A$ to MATH. The key to the proof is to use compute $AD$ using MATH and $AC$, compute $AD$ using MATH and $AB$, set the results equal to one another, and then simplify. Notice that there are a number of cases that need to be considered: $B-C-D,$ $C-D-B,$ $D=B,$ $D=C$ and $D-B-C$.


a. Prove the result when $C-D-B.$ Include a diagram that illustrates this case. (4 pts.)

First note that, in this case, $\angle C$ and $\angle B$ are acute angles. To see this, note that since $C-D-B$ we have that MATH and hence MATH As $\Delta ACD$ is a right triangle with right angle at $D,$ it follows that $\angle ACD$ and hence $\angle ACB$ is acute. A similar argument shows that $\angle ABC$ is acute.

Now
MATH
and the result follows.


b. Prove the result when $C-B-D.$ Include a diagram that illustrates this case.(4 pts.)


In this case $\angle C$ is acute and $\angle B$ is obtuse. To see this note that $C-B-D$ yields that MATH and hence MATH As $\Delta ACD$ is a right triangle with right angle at $D,$ it follows that $\angle ACD$ and hence $\angle ACB$ is acute. Now note that $C-B-D$ yields that MATH and MATH are opposite rays; hence $\angle ABD$ and $\angle ABC$ are a linear pair and hence supplementary. As $\Delta ACD$ is a right triangle with right angle at $D,$ it follows that $\angle ABD$ is acute and hence $\angle ABC$ is obtuse.

Now
MATH
and, as MATH by definition,


MATH
and the result follows.

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