Section 3.4
Beyond CPCTC
Gabby Shefski
Objectives
Identify medians of triangles
Identify altitudes of triangles
Understand why auxiliary lines are used in
some proofs
Write proofs involving steps beyond
CPCTC
Medians of Triangles
Definition: A median of a triangle is a line
segment drawn from any vertex of the
triangle to the midpoint of the opposite
side. (A median divides into two congruent
segments, or bisects the side to which it is
drawn.)
Every triangle has three medians.
The point at which all three medians
intersect is the centroid.
Samples
A
P
E
F
D
B
||
D
||
AD, CE, and BF are
medians of ∆ABC.
C
Q
QD is a median of ∆ABC.
R
Altitudes of Triangles
Definition: An altitude of a triangle is a line
segment drawn from any vertex of the
triangle to the opposite side, extended if
necessary, and perpendicular to that side.
(An altitude of a triangle forms right angles
with one of the sides.)
Every triangle has three altitudes.
The point at which all three altitudes
intersect is the orthocenter.
Samples
A
H
I
F
B
D
AD and BF are
altitudes of ∆ABC.
C
J
HI and JI are
altitudes of ∆HIJ.
Auxiliary Lines
Definition: Auxiliary lines are additional
lines, segments, or rays added to a
diagram that do not appear in the original
figure. They can connect two points that
are already present in the figure.
Postulate: Two points determine a line (or
ray or segment)
Steps Beyond CPCTC
After using CPCTC to prove angles or
segments congruent, you can now find
altitudes, medians, angle bisectors,
midpoints, etc.
Sample Problem
Given: AD is an altitude and
a median of ∆ABC
Prove: AB ≈ BC
A
B
D
C
Solution
Statements
1. AD is an altitude
and median of
∆ABC
2. <ADB, <ADC
are rt. <s
3. <ADB ≈ <ADC
4. BD ≈ CD
5. AD ≈ AD
6. ∆ ABD ≈ ∆ ADC
7. AB ≈ AC
Reasons
1. Given
2. An altitude of a
triangle forms rt
<s with the side
to which it is
drawn.
3. If two <s are rt.
<s, then they are
≈
4. A median of a
triangle divides
the side to which
it is drawn into 2
≈ segments.
5. Reflexive
6. SAS (3, 4, 5)
7. CPCTC
A
B
D
C
Sample Problem
Given: AB ≈ AC
<ABD ≈ <CBD
Prove: AD bisects BC
A
B
D
C
Solution
Statements
1.
AB ≈ AC
2.
<BAD ≈ <CAD
3.
AD ≈ AD
4.
∆ ABD ≈ ∆ ACD
5. BD ≈ DC
6.
AD bisects BC
Reasons
1.
Given
2.
Given
3.
Reflexive
4.
SAS (1, 2, 3)
5. CPCTC
6.
If a segment
divides another
seg. into 2 ≈ segs,
then it bisects the
segment
A
B
D
C
Practice Problem
Given: <E ≈ <G
<ABF ≈ <ADF
EB ≈ GD
Prove: AF bisects EG
A
B
E
C
D
F
G
Solution
Statements
1.
<E ≈ <G
2.
<ABF ≈ <ADF
3.
<ABF suppl.
<FBE
4.
<ADF suppl.
<FDG
5.
<FBE ≈ <FDG
6.
EB ≈ GD
7.
∆ EBF ≈ ∆
GDF
8.
EF ≈ FG
9.
AF bisects EG
Reasons
1.
Given
2.
Given
3.
If two <s form
a st. < then
they are suppl.
4.
Same as 3
5.
Suppl. of ≈ <s
are ≈
6.
ASA (1, 6, 5)
7.
CPCTC
8.
If a segment
divides
another
segment into 2
≈ segments,
then it bisects
the segment
A
B
E
C
D
F
G
Works Cited
Rhoad, Richard, George Milauskas, and
Robert Whipple. Geometry for Enjoyment
and Challenge. Evanston, IL: McDougal,
Littell, 1991. Print.