1. Section 3.4 Beyond CPCTC
Gabby Shefski

2. Objectives Identify medians of triangles
Identify altitudes of triangles
Understand why auxiliary lines are used in some proofs
Write proofs involving steps beyond CPCTC

3. 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.

4. Samples AD, CE, and BF are medians of ∆ABC. QD is a median of ∆ABC. A
B || D || C Q R
AD, CE, and BF are
medians of ∆ABC.
QD is a median of ∆ABC.

5. 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.

6. Samples AD and BF are altitudes of ∆ABC.
H I F
B D C J
AD and BF are altitudes of ∆ABC.
HI and JI are altitudes of ∆HIJ.

7. 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)

8. Steps Beyond CPCTC
After using CPCTC to prove angles or segments congruent, you can now find altitudes, medians, angle bisectors, midpoints, etc.

9. Sample Problem
Given: AD is an altitude and a median of ∆ABC Prove: AB ≈ BC A
B D C

10. Solution Statements AD is an altitude and median of ∆ABC
<ADB, <ADC are rt. <s
<ADB ≈ <ADC
BD ≈ CD
AD ≈ AD
∆ ABD ≈ ∆ ADC
AB ≈ AC
Reasons
Given
An altitude of a triangle forms rt <s with the side to which it is drawn.
If two <s are rt. <s, then they are ≈
A median of a triangle divides the side to which it is drawn into 2 ≈ segments.
Reflexive
SAS (3, 4, 5)
CPCTC A
B D C

11. Sample Problem Given: AB ≈ AC <ABD ≈ <CBD Prove: AD bisects BC A

12. Solution Statements AB ≈ AC <BAD ≈ <CAD AD ≈ AD ∆ ABD ≈ ∆ ACD
BD ≈ DC
AD bisects BC
Reasons
Given
Reflexive
SAS (1, 2, 3)
CPCTC
If a segment divides another seg. into 2 ≈ segs, then it bisects the segment A
B D C

13. Practice Problem
Given: <E ≈ <G <ABF ≈ <ADF EB ≈ GD Prove: AF bisects EG A
B C D
E F G

14. Solution Statements <E ≈ <G <ABF ≈ <ADF
<ABF suppl. <FBE
<ADF suppl. <FDG
<FBE ≈ <FDG
EB ≈ GD
∆ EBF ≈ ∆ GDF
EF ≈ FG
AF bisects EG
Reasons
Given
If two <s form a st. < then they are suppl.
Same as 3
Suppl. of ≈ <s are ≈
ASA (1, 6, 5)
CPCTC
If a segment divides another segment into 2 ≈ segments, then it bisects the segment A
B C D
E F G

15. Works Cited
Rhoad, Richard, George Milauskas, and Robert Whipple. Geometry for Enjoyment and Challenge. Evanston, IL: McDougal, Littell, Print.