* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download geometry module 1 lesson 29 special lines in
Survey
Document related concepts
Penrose tiling wikipedia , lookup
History of geometry wikipedia , lookup
Multilateration wikipedia , lookup
Technical drawing wikipedia , lookup
Dessin d'enfant wikipedia , lookup
Golden ratio wikipedia , lookup
Apollonian network wikipedia , lookup
Euler angles wikipedia , lookup
Line (geometry) wikipedia , lookup
Rational trigonometry wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Trigonometric functions wikipedia , lookup
History of trigonometry wikipedia , lookup
Euclidean geometry wikipedia , lookup
Transcript
GEOMETRY MODULE 1 LESSON 29 SPECIAL LINES IN TRIANLGES OPENING EXERCISE A. Use your compass and straightedge to find the midpoints of Μ Μ Μ Μ π΄π΅ and Μ Μ Μ Μ π΄πΆ . Label the midpoint as X and Y, respectively. Μ Μ Μ Μ . B. Draw the midsegment ππ The midsegment is a line segment that joins the midpoints of two sides of a triangle or trapezoid. ο· Compare β π΄ππ and β π΄π΅πΆ. Compare β π΄ππ and β π΄πΆπ΅. Without using a protractor, what would you guess is the relationship between these two pairs of angles? πβ π΄ππ = πβ π΄π΅πΆ πβ π΄ππ = πβ π΄πΆπ΅ ο· Μ Μ Μ Μ ? From your conclusion above, what can now be said about Μ Μ Μ Μ ππ and π΅πΆ Μ Μ Μ Μ Μ Μ Μ Μ β₯ π΅πΆ ππ HOMEWORK QUESTIONS Problem Set Module 1 Lesson 28 is due Thursday. MOD1 L29 1 DISCUSSION Μ Μ Μ Μ and π΄πΆ Μ Μ Μ Μ , we could have chosen any Even though we chose to determine the midsegment of π΄π΅ two sides to work with. Letβs now focus on the properties associated with a midsegment. The midsegment of a triangle is parallel to the third side of the triangle and half the length of the third side of the triangle. Given: Μ Μ Μ Μ ππ is a midsegment of βπ΄π΅πΆ. Μ Μ Μ Μ and ππ = 1 π΅πΆ Μ Μ Μ Μ β₯ π΅πΆ Prove: ππ 2 Μ Μ Μ Μ to point G so that ππ = ππΊ. Draw πΊπΆ. Construction: Extend ππ STEP JUSTIFICATION 1 Μ Μ Μ Μ is a midsegment of βπ΄π΅πΆ ππ Given 2 ππ = ππΊ Construction 3 πβ π΄ππ = πβ πΊππΆ Vertical Angles 4 π΄π = ππΆ Property of Midpoint 5 βπ΄ππ β βπΆπΊπ SAS 6 πΊπΆ = π΄π Corresponding sides of congruent triangles are equal in length. 7 πΊπΆ = π΅π Substitution (Line 1: π΄π = π΅π) 8 πβ π΄ππ = πβ πΆπΊπ Corresponding angles of congruent triangles are equal in measure. 9 Μ Μ Μ Μ π΄π΅ β₯ Μ Μ Μ Μ πΊπΆ Alternate Interior Angles Converse 10 BXGC is a parallelogram 11 12 ππΊ = π΅πΆ Μ Μ Μ Μ Μ Μ Μ Μ ππ β₯ π΅πΆ ππΊ = 2ππ Substitution (Line #2 and ππ + ππΊ = ππΊ) 13 π΅πΆ = 2ππ Substitution (Line #11) 14 MOD1 L29 ππ = 1 π΅πΆ 2 One pair of opposite sides is equal and parallel. (Line 7 and 9) Property of Parallelogram Division 2 PRACTICE Apply what you know about the properties of midsegments to solve the following exercises. 1. a. Find the length of x. 15 b. Find the perimeter of βπ΄π΅πΆ. 68 2. a. Find the angle measure of x. 50° b. Find the angle measure of y. 70° 3. In βπ ππ, the midpoints of each side have been marked by points X, Y, and Z. ο· Mark the halves of each side divided by the midpoint with a congruency mark. ο· Draw midsegments Μ Μ Μ Μ ππ, Μ Μ Μ Μ ππ, and Μ Μ Μ Μ ππ. March each midsegment with the appropriate congruency mark for the sides of the triangle. a. What conclusion can you draw about the four triangles within βπ ππ? Why? All four triangles are congruent by SSS. b. State the appropriate correspondences among the four triangles within βπ ππ. βπ ππ, βπππ, βπππ, βπππ c. State a correspondence between βπ ππ and nay one of the four small triangles. βπ ππ, βπ ππ MOD1 L29 3 ON YOUR OWN 1. Find x. Μ Μ Μ Μ Μ and Μ Μ Μ Μ Μ 2. R and S are the midpoints of ππ ππ, respectively. Find the perimeter of βπππ. SUMMARY The midsegment of a triangle is parallel to the third side of the triangle and half the length of the third side of the triangle. 1. 79° 2. 82 MOD1 L29 4