Medians and Altitudes of

TrianglesSection 3.4

Heidi Frantz, T.J. Murray

Median of a Triangle

A. Line segment drawn from triangle vertex to midpoint of opposite side. [Ex. (segment) FB.]

B. Line segment drawn from triangle vertex that bisects opposite side.

C. Line segment drawn from triangle vertex that divides opposite side into 2 congruent segments (used in proofs).

SAMPLE MEDIAN PROBLEMS

1. Given: CE congruent EBConclusion: AE is median to CBReason: If a segment drawn from a

triangle vertex divides the opposite side into 2 congruent segments, then it is a median.

2. Given: BF is median to ACConclusion: FC congruent FAReason: If a segment drawn from a triangle vertex is a median, then it divides the opposite side into two congruent segments.

CENTROID AND ORTHOCENTER I. Centroid

A. Center of gravity of triangleB. Two-thirds of the way from the vertex to the midpoint of the triangle.

II. OrthocenterA. Where all 3 altitudes of a triangle intersectB. One of triangle's points of concurrency

Altitudes Of Triangles

A. Line Segment drawn from triangle vertex perpendicular to opposite side (extended if necessary; proofs).

B. Line segment drawn from a triangle vertex that forms right angles with the opposite side (Proofs).C. Line segment drawn from triangle vertex that forms 90 degree angles with the opposite side (Problems).

Every Triangle has 3 altitudes

SAMPLE ALTITUDE PROBLEMS

1. Given: AD is perpendicular to BC

Conclusion: AD is alt. to BCReason: If a segment drawn

from a triangle vertex is perpendicular to the opposite side, then it is an altitude.

SAMPLE ALTITUDE PROBLEMS

2. Given: AD is alt. of triangle ABC

Conclusion: Angle ADC is a right angle

Reason: If a segment drawn from a triangle vertex is an altitude , then it forms right angles with the opposite side.

MEDIAN, ALTITUDE PRACTICE PROBLEMS

Given: Triangle ABC is isosceles with base BC

AD is alt. of triangle ABC

Prove: AD is median of triangle ABC

NOTE: DIAGRAM NOT DRAWN TO SCALE

1.

ANSWERS

1. Triangle ABC is isos. w/ base BC (Given)2. AD is alt. of Triangle ABC (Given)3. AB=AC (If triangle is isos., then sides are congruent.)4. Angle ADB, ADC are right angles (If a segment drawn from a

triangle vertex is an alt., then it forms right angles with the opposite side.)

5. Triangle ADB, ADC are right triangles (If a triangle contains a right angle, then it is a right triangle.)

6.AD=AD (Reflexive)7. Triangle ADB=ADC (HL, steps 3,5,6)8. BD=DC (CPCTC)9. AD is median of ABC (If a segment drawn from a triangle

vertex divides the opposite side into congruent segments, then it is a median.

Statements are numbered, and reasons are in parenthesis.

MEDIAN PRACTICE PROBLEMS

Given: AE, FB, and DC are medians.AF=10, AB=45, CE=x+10, EB=2x-10

Find the perimeter of triangle ABC

2.

ANSWERS

AF=FC (10+10=20) (CE=EB) x+10=2x-10

AC=20, AB=45 20=xx+10=30, 2x-

10=30CB=

(30+30)CB=60

AB+AC+CB= (20+45+60) = 125Therefore, perimeter of triangle ABC is

125 units.

ALTITUDE PRACTICE PROBLEMS

Given: AD is alt. to triangle ABC

Angle BDA=6x Angle BAD=x Angle DAC=3x+10

Find: Measure of angle BAC

3.

ANSWER

m<BDA=90 degrees, 6x (90=6x) (x=15)

m<BAD=x, which is = to 15 degrees. m<DAC=3x+10, which is = to 55

degrees.(m<BAD) + (m<DAC) = (m<BAC) 15 degrees + 55 degrees= 70 degreesTherefore, m<BAC=70 degrees

ALTITUDE PRACTICE PROBLEMS

Given: Angle BDA is a right angle

Conclusion: ________

Reason:_____________________________________________________________________________

ANSWER

Conclusion: AD is altitude to triangle ABCReason: If a line segment drawn from a

triangle vertex forms 90 degree angles with the opposite side, then it is an altitude to that side.

Works Cited

Rhoad, Richard, George Milauskas, and Robert Whipple. Geometry for Enjoyment and Challenge. New ed. Evanston, Illinois: McDougal, Little, and Company, 2004. 131-137. Print.

"File:Triangle.Centroid.Median.png." WIKIMEDIA COMMISIONS. 18 January 2009. Web. 17 Jan 2010. <http://commons.wikimedia.org/wiki/File:Triangle.Centroid.Median.png>.