scribe notes for december 7, 2009 similarities we began class

8/14/2019 Scribe Notes for December 7, 2009 Similarities We Began Class

1/5

Scribe Notes for December 7, 2009Similarities

We began class by talking about similar triangles. We had a homework assignment from lastweek that involved similar triangles and theorems involving similarities. Today we went over

most of that homework assignment.

First we began with the definition of similar triangles:Two triangles ABC and ABC are said to be similar , denoted by ABC ~ ABC if < A< A, < B< B, and < C < C.

We then talked about the Fundamental Theorem of Similar Triangles . This theorem statesthat two triangles ABC and ABC are similar if and only if their corresponding sides areproportional, i.e.

(AB)

=

( )

=( )

(

)

Proof:In proving this theorem we have to go both ways because it is an if and only if theorem.

We begin with ABC ~ ABC and want to prove that (AB)

=( )

=( )

(

). In our proof

we will let ABC be the smaller triangle without loss of generality.

We copy segment AB onto segment AB such that we AB is congruent to AB. We thencopy angle B to angle B so that we create point C on AC. We know that C will be betweenpoints A and C because if AB < AB then AC < AC. We did not prove this in class buttalked about being able to prove it by contradiction.


2/5

We can extend line BC so that we create a new point E, such that B*C*E. Now, by verticalangles we see that < ECC < ACB. Since < C< C we know that lines BC and BC areparallel (by Alternate Interior Angles).

Now we can use the Parallel Projection Theorem to say that(AC")( ")

=( " )( " )

. If we take the

reciprocal of both sides and set them equal to each other (we know they will in fact still equaleach other), add one to both sides, and then combine the fractions by getting a commondenominator we get that

+ ( " )( ") =

( B)+ (B )( ") . From here we easily conclude that

( )

=

(AB)

.

We know that we can repeat this process but copy the segment AB (or another segment fromABC) differently so that we conclude that a different proportion holds. From this we canthen conclude that

(AB)

=

( )

=( )

(

).

Now, we have to prove this theorem the other way. Here we start with(AB)

=

( )

=( )

(

)

and want to prove that ABC ~ ABC. We begin with ABC and ABC.

We then copy AB onto AB as before. This time, we will create a parallel line that intersectsAC at a point C from our point B. So BC is parallel to BC. We now have that ACB ~ ACB by our construction.


3/5

Now, because of our similar triangles we know

(AB")( ) =

( ")( ) =

( " ")

( ). The hypothesis tells us

that(A B )( ) =

( )( ) =

( )

( ). This implies that

( ")( ) =

( )( ) . Which implies that ( ") =

( ) and ( C') = ( " ") . Then we see by SSS that C . Thus ABC ~ ABC.

We then talked about SAS Similarity Criterion . This states that if < A < A and(AB)

=

( )

then ABC ~ ABC.

Proof:We began with ABC and ABC. We then constructed a new triangle in ABC, such

that it is similar to ABC, with one of its sides congruent to one of the sides of ABC. Wethen used proportions to show that the new triangle is congruent to ABC, and thereforeABC ~ ABC. This picture is the same as in the above picture so I have not included a newone, and the proportions used are similar as well.

Next we talked about the Converse of the Parallel Projection Theorem . We wanted to know if it was true, sometimes true, or false. The converse of PPT is: given A*B*C and A*B*C and if (AB)

( )=

( )(

)then is parallel to which is parallel to . We decided that this is

always true only when either point A=A or point C=C. We can see that this will create similartriangles and will therefore hold. However, if this does not occur it is not guaranteed that the

lines will be parallel.

We then moved on to the Midpoint Theorem . This theorem was easily proven usingsimilarities, however when we tried to prove it without using similarities we struggled (howeverwe did come to a conclusion eventually). The Midpoint Theorem states that given ABC, if weconnect the midpoints of two sides we will have a new segment that is parallel to the third side of the triangle.


4/5

Here we have that < DCE < ACB. We know that

(CE)( )

=( )( ) . By the hypotheses of SAS for

similar triangles we know that CDE ~ CAB. Then we know that since these triangles aresimilar, their corresponding angles are congruent. Thus, we can use what we know aboutcongruent angles created in this fashion (with a transversal through two parallel lines) to knowthat DE is parallel to AB.

To prove the midpoint theorem without using similarities we first let E be the midpoint of BC.We will let D be the point such that A*E*D. From this we get AEC DEB (by SAS) and EDAE. From this we get that AC and BD are parallel and congruent. Ralph then came up with atheorem that says that this implies that ABDC is parallelogram. He told us that he would provethis theorem for us next time. Next, we will draw a parallel to AB through E, and let it intersectAC at point F and BD at point G. Then FGDC (or FGBA) is a parallelogram, so FC GD(FABG). We also know that CEF BEG by ASA, so FC BG, and now that tells you thatFCAF. So from this we see that F is the midpoint of AC and E is the midpoint of BC.


5/5

scribe notes for december 7, 2009 similarities we began class

Documents