Section 3.4Beyond CPCTC

Gabby Shefski

Objectives

Identify medians of trianglesIdentify altitudes of trianglesUnderstand why auxiliary lines are used in

some proofsWrite 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

B || D || C Q R

AD, CE, and BF aremedians of ∆ABC.

E F

P

D

QD is a median of ∆ABC.

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

B D C J

H I

AD and BF are altitudes of ∆ABC.

F

HI and JI are altitudes of ∆HIJ.

A

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

A

B D C

Statements1. AD is an altitude

and median of ∆ABC

2. <ADB, <ADC are rt. <s

3. <ADB ≈ <ADC4. BD ≈ CD5. AD ≈ AD6. ∆ ABD ≈ ∆ ADC7. AB ≈ AC

Reasons1. Given2. 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. Reflexive6. SAS (3, 4, 5)7. CPCTC

Sample Problem

Given: AB ≈ AC

<ABD ≈ <CBD

Prove: AD bisects BC

A

B D C

Solution

A

B D C

Statements1. AB ≈ AC2. <BAD ≈ <CAD3. AD ≈ AD4. ∆ ABD ≈ ∆ ACD5. BD ≈ DC6. AD bisects BC

Reasons1. Given2. Given3. Reflexive4. SAS (1, 2, 3)5. CPCTC6. If a segment

divides another seg. into 2 ≈ segs, then it bisects the segment

Practice Problem

Given: <E ≈ <G

<ABF ≈ <ADF

EB ≈ GD

Prove: AF bisects EG

A

B C D

E F G

Solution

A

B C D

E F G

Statements1. <E ≈ <G2. <ABF ≈ <ADF3. <ABF suppl.

<FBE4. <ADF suppl.

<FDG5. <FBE ≈ <FDG6. EB ≈ GD7. ∆ EBF ≈ ∆

GDF8. EF ≈ FG9. AF bisects EG

Reasons1. Given2. Given3. If two <s form

a st. < then they are suppl.

4. Same as 35. Suppl. of ≈ <s

are ≈ 6. ASA (1, 6, 5)7. CPCTC8. If a segment

divides another segment into 2 ≈ segments, then it bisects the segment

Works Cited

Rhoad, Richard, George Milauskas, and Robert Whipple. Geometry for Enjoyment and Challenge. Evanston, IL: McDougal, Littell, 1991. Print.