Capítulo 4. Base teórica
4.3 Análisis de la respuesta en el dominio temporal
4.3.4. Transformada discreta de Fourier
We’re finally ready to prove a few theorems. In every case, the proofs are given in verbal form, and also as statements/reasons (S/R) tables. The theorems in this section are stated as “problems,” and the proofs are stated as “solutions.”
PROBLEM 5-5
Let P, Q, R, and S be mutually distinct points, all of which lie in the same plane. Suppose lines PQ, QR, RS, and SPare all mutually distinct (that is, no two of them coincide). Consider line segments PQ, QR, RS, and SP, each of which lies on the line having the same name. The line segments form a four-sided figure PQRS, with vertices at points P, Q,
R, and S, in that order proceeding counterclockwise. Suppose the fol- lowing are true:
•
Line segment PQhas the same length as line segment SR•
Line segment SPhas the same length as line segment RQConsider the line segment SQ, which divides the figure PQRSinto two triangles, ∆SPQand ∆QRS. Prove that ∆SPQ≡ ∆QRS.
SOLUTION 5-5
It helps to draw a diagram of this situation. If you wish, you can do this based on the description of the problem, and get something like Fig. 5-13. First, let’s decide upon (that is, assign) corresponding sides for the tri- angles, proceeding counterclockwise around the triangle in either case:
•
Line segment SPin ∆SPQcorresponds to line segment QRin ∆QRS•
Line segment PQin ∆SPQcorresponds to line segment RSin ∆QRSWe are told that line segment PQ has the same length as line seg- ment SR. The length of a line segment does not depend on the direction in which it is expressed. Therefore, line segment PQin ∆SPQhas the same length as line segment RSin ∆QRS. These line segments are cor- responding sides in the triangles.
We are also told that line segment SP has the same length as line segment RQ. Therefore, line segment SPin ∆SPQhas the same length as line segment QR in ∆QRS. These line segments are corresponding sides in the triangles.
Finally, from the definition of the length of a line segment, we know that line segment QSin ∆SPQhas the same length as line segment SQ
in ∆QRS. This line segment constitutes corresponding sides (which also happen to coincide) in the triangles.
We have shown that the corresponding sides of ∆SPQ and ∆QRS
have identical lengths when we proceed around the triangles in the same direction, counterclockwise in this case. Therefore, according to the SSS axiom, we can conclude that that ∆SPQ≡ ∆QRS. Table 5-1 is an S/R version of this proof.
PROBLEM 5-6
Let P, Q, R, and S be mutually distinct points, all of which lie in the same plane. Suppose lines PQ, QR, RS, and SPare all mutually dis- tinct. Consider line segments PQ, QR, RS, and SP, each of which lies on the line having the same name. This forms a four-sided figure
PQRS, with vertices at points P, Q, R, and S, in that order proceeding counterclockwise. Suppose the following are true:
•
Line segment PQ has the same length as line segment SR•
Line segment SPhas the same length as line segment RQP Q
R S
Imagine lines PQand SR, the extensions of line segments PQand SR, respectively. Also imagine line SQ, the extension of line segment SQ. Prove that m∠QSR= m∠SQP.
SOLUTION 5-6
If you wish, you can draw a diagram of this situation based on the pre- vious problem and on the description of this problem, and get some- thing like Fig. 5-14.
Problem 5-5, now that it has been solved, is a theorem. In the current problem, therefore, we know that ∆SPQ≡ ∆QRS. When two triangles are
Statements Reasons
Line segment SPin ∆SPQ We assign them that way. corresponds to line segment QR
in ∆QRS.
Line segment PQin ∆SPQ We assign them that way. corresponds to line segment RS
in ∆QRS
Line segment QSin ∆SPQ We assign them that way. corresponds to line segment SQ
in ∆QRS.
Line segment PQ has the same Given. length as line segment SR.
Line segment SPhas the same Given. length as line segment RQ.
Line segment QS has the same This comes from the definition length as line segment SQ. of the length of a line segment: it is the same in either direction.
Corresponding sides of ∆SPQand This is based on the above state-
∆QRS have the identical lengths ments, and on the way we have expressed counterclockwise. assigned corresponding sides.
∆SPQ ≡ ∆QRS. SSS axiom.
Table 5-1. An S/R version of the proof demonstrated in Solution 5-5 and
directly congruent, then the counterclockwise measures of their corre- sponding angles, as we proceed in the same direction around either tri- angle, are equal. From the description of this situation (and with the help of the drawing, if you need it), it is evident that ∠QSRand ∠SQPare corresponding angles, the first angle in ∆QRS, and the second angle in
∆SPQ. The angle measures, as specified in the statement of this problem, are expressed counterclockwise. Therefore, their measures are equal; that is, m∠QSR= m∠SQP. Table 5-2 is an S/R version of this proof.
P Q
R S
Fig. 5-14. Illustration for Problem 5-6.
Statements Reasons
∆SPQ≡ ∆QRS This is the theorem resulting from Solution 5-5.
Counterclockwise measures of This comes from the definition of corresponding angles in ∆QRS direct congruence.
and ∆SPQare equal.
∠QSRand ∠SQPare corresponding This is evident from examination angles in ∆QRSand ∆SPQ, as we of the problem.
proceed in the same direction around both triangles.
m∠QSRand m∠SQPare This is evident from the statement defined counterclockwise. of the problem.
m∠QSR = m∠SQP This comes from information derived in the preceding steps, and from the definition of direct congruence.
Table 5-2. An S/R version of the proof demonstrated in Solution 5-6 and
PROBLEM 5-7
In the situation of the previous problem, let T be a point on line PQ
such that point Qis between point Pand point T. Let Ube a point on line SRsuch that point Sis between point Uand point R. Suppose that, in addition to the other conditions in the previous problem, the follow- ing are true:
•
Line segment QThas the same length as line segment US•
Line segment UQhas the same length as line segment STProve that m∠USQ= m∠TQS.
SOLUTION/EXERCISE 5-7
Try this for yourself! Here are some hints:
•
Construct triangles ∆UQSand ∆TSQ•
Show that ∆UQSand ∆TSQare directly congruent•
Use the same approach as in Solution 5-6•
Feel free to use Fig. 5-15 as a visual aidIf you have trouble proving this, accept it on faith for now, and come back to it tomorrow. We have all had the experience of solving a tough problem that turned out to be easy after we “slept on it.” This technique often works well with elusive math proofs.