1 / 16

Medians and Altitudes of Triangles

Medians and Altitudes of Triangles. Section 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.

oscar-workman
Download Presentation

Medians and Altitudes of Triangles

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Medians and Altitudes of Triangles Section 3.4 Heidi Frantz, T.J. Murray

  2. 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).

  3. 1. Given: CE congruent EB Conclusion: AE is median to CB Reason: If a segment drawn from a triangle vertex divides the opposite side into 2 congruent segments, then it is a median. Sample median problems 2. Given: BF is median to AC Conclusion: FC congruent FA Reason: If a segment drawn from a triangle vertex is a median, then it divides the opposite side into two congruent segments.

  4. Centroid and Orthocenter • I. Centroid A. Center of gravity of triangle B. Two-thirds of the way from the vertex to the midpoint of the triangle. II. Orthocenter A. Where all 3 altitudes of a triangle intersect B. One of triangle's points of concurrency

  5. Altitudes Of Triangles A. Line Segment drawn from triangle vertex perpendicular to opposite side (extended if necessary; proofs). • Every Triangle has 3 altitudes 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).

  6. 1. Given: AD is perpendicular to BC Conclusion: AD is alt. to BC Reason: If a segment drawn from a triangle vertex is perpendicular to the opposite side, then it is an altitude. Sample altitude problems

  7. 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. Sample altitude problems

  8. 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 Median, Altitude practice problems 1.

  9. 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. Answers Statements are numbered, and reasons are in parenthesis.

  10. Given: AE, FB, and DC are medians. AF=10, AB=45, CE=x+10, EB=2x-10 Find the perimeter of triangle ABC Median practice problems 2.

  11. AF=FC (10+10=20) (CE=EB) x+10=2x-10 AC=20, AB=45 20=x x+10=30, 2x-10=30 CB= (30+30) CB=60 AB+AC+CB= (20+45+60) = 125 Therefore, perimeter of triangle ABC is 125 units. Answers

  12. Given: AD is alt. to triangle ABC Angle BDA=6x Angle BAD=x Angle DAC=3x+10 Find: Measure of angle BAC Altitude practice problems 3.

  13. 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 degrees Therefore, m<BAC=70 degrees Answer

  14. Given: Angle BDA is a right angle Conclusion: ________ Reason:_____________________________________________________________________________ Altitude practice problems

  15. Conclusion: AD is altitude to triangle ABC Reason: 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. answer

  16. 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>.

More Related