Transcript Parallel Lines and Proportional Parts

Parallel Lines and
Proportional Parts
By: Jacob Begay
Theorem 7-4 Triangle
Proportionality:

If a line is parallel to one side of a
triangle and intersects the other two
sides in two distinct points, then it
separates these sides into segments
of proportional lengths.
C
C
B
A
D
E
C
B
CB
CA
=
D
CD
CE
=
BD
AE
A
E
Theorem 7-5 Converse of the
Triangle Proportionality:

If a line intersects two sides of a
triangle and separates the sides into
corresponding segments of
proportional lengths, then the line is
parallel to the third side.
C
BD
D
B
A
E
AE
Theorem 7-6 Triangle Midpoint
Proportionality:

A segment whose endpoints are the
midpoints of two sides of a triangle is
parallel to the third side of the
triangle, and its length is one-half
the length of the third side.
C
B
D
BD ll AE
A
E
2BD=AE OR BD=1/2AE
Corollary 7-1

If three or more parallel lines
intersect two transversals, then they
cut off the transversals
proportionally.
D
B
C
AB = AD
AE AG
A
E
BC CD
=
EF FG
F
G
AC BC
=
AF EF
CD FG
=
AE AB
Corollary 7-2

If three or more parallel lines cut off
congruent segments on one
transversal, then they cut off
congruent segments on every
transversal.
B
E
C
F
D
BE
G
CF
GD
Example

Based on the figure below, which
statement is false?
A
4
3
E
D
3
4
B
C
A.DE is Parallel to BC
C.ABC ~ ADE
B.D is the Midpoint of AB
D.ABC is congruent to ADE
D. ABC is congruent to ADE. Corresponding sides of the triangles
are proportional but not congruent.
Example

Find the value of X so that PQ is
parallel to BC.
A
3
P
X+0.25
A.1
B
B.2.5
4
Q
3
C
C.1.25
D.2
D. 2 Since the corresponding segments must be proportional for
PQ to be parallel to BC.
Example




Triangle ABC has vertices A (0,2), B (12,0), and C (2,10).
A. Find the coordinates of D, the midpoint of Segment AB, and E, the
midpoint of Segment CB.
B. Show that DE ll AC.
C. Show that 2DE = AC.
D=
E=
0+12, 2+0
2
Or D = (6,1)
2
12+2, 0+10
Midpoint Segment CB (7,5)
Or E = (7,5)
2
2
Slope of AC = 2-10
AC=4
0-2
2
AC ll DE
2
AC= (0-2) + (2-10)
=
4+64
=
68
Midpoint Segment AB (6,1)
Slope of DE = 1-5
DE=4
DE =
=
Or 2
17
6-7
2
2
(6-7) + (1-5)
1+16 Or
Therefore 2DE = AC
17