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SPECIAL SEGMENTS IN A TRIANGLE MEDIANS ARE PROPORTIONAL PROBLEM 6

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Presentation on theme: "SPECIAL SEGMENTS IN A TRIANGLE MEDIANS ARE PROPORTIONAL PROBLEM 6"— Presentation transcript:

1 SPECIAL SEGMENTS IN A TRIANGLE MEDIANS ARE PROPORTIONAL PROBLEM 6
STANDARDS 4 and 5 MEDIANS ARE PROPORTIONAL PROBLEM 6 ALTITUDES ARE PROPORTIONAL ANGLE BISECTORS ARE PROPORTIONAL PROBLEM 7 PERIMETERS ARE PROPORTIONAL PROBLEM 8 ANGLE BISECTORS FORM PROPOR. SEG. PROBLEM 9 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 Students prove basic theorems involving congruence and similarity.
Standard 4: Students prove basic theorems involving congruence and similarity. Los estudiantes prueban teoremas básicos que involucran congruencia y semejanza. Standard 5: Students prove triangles are congruent or similar and are able to use the concept of corresponding parts of congruent triangles. Los estudiantes prueban que triángulos son congruentes o semejantes y son capaces de usar el concepto de partes correspondientes de triángulos congruentes. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 ALTITUDE: A segment from a vertex of a triangle perpendicular to the line containing the opposite side. ALTURA: Un segmento desde el vértice de un triángulo perpendicular a la línea conteniendo el lado opuesto. STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 PERPENDICULAR BISECTOR:
A line or segment that passes through the midpoint of a side of a triangle and is perpendicular to that side.. BISECTRIZ PERPENDICULAR: Una línea o segmento que pasa a través del punto medio de un lado de un triángulo y es perpendicular a ese lado. STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 MEDIAN: A segment that connects a vertex of a triangle to the midpoint of the opposite side. MEDIANA: Un segmento que conecta un vertice de un triángulo a el punto medio del lado opuesto. STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 ANGLE BISECTOR: A segment from a vertex to the opposite side that bisects (divides in two equal parts) the angle of the triangle. BISECTRIZ ANGULAR: Un segmento desde un vértice a el lado opuesto que biseca (divide en dos partes iguales) el ángulo de el triángulo. STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe? Altitude STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 Perpendicualr Bisector
Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe? Perpendicualr Bisector STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe? Median STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe? Angle Bisector STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 Lets put them all together
Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe? Lets put them all together Altitude Perpendicualr Bisector Median Angle Bisector So, all of them occupy the same Geometric Space in the triangle! Could you do something similar with an equilateral triangle in all the vertices? STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 REVIEW: CAB MKL A CA AB BC B C LM MK KL KL AB MK CA LM BC K M L OR
STANDARDS 4 and 5 REVIEW: CAB MKL A CA AB BC B C LM MK KL KL AB MK CA LM BC OR K M L PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13 In simlar triangles MEDIANS are proportional to sides:
AD AB KL CA MK BC LM B C KN D K KL AB MK CA LM BC KN AD OR M N L STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14 IJK TJU, R is the midpoint of TU and S is the midpoint of IK
IJK TJU, R is the midpoint of TU and S is the midpoint of IK. TU= 10, JR= 6, IK = 30. Find JS. JS JR TU IK J I K S U T R S R J U T I K (6) (6) JS 6 10 30 = = = (30)(6) 10 JS = 180 10 JS JS=18 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15 In similar triangles ALTITUDES are proportional to sides:
AE AB KL CA MK BC LM B C O KO K KL AB MK CA LM BC KO AE OR M L STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16 In similar triangles ANGLE BISECTORS are proportional to sides:
F AF AB KL CA MK BC LM B C P KP K KL AB MK CA LM BC KP AF OR M L STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17 JS is an angle bisector. If TJ = 14, IT=22, and JS= 30, what is the value for JR? Suppose IJK TJU.
36 14 22 JR JT = (30) (30) JR 30 36 14 = JS JT+TI = (14)(30) 36 JR = 420 36 JR JR = 11.7 . STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18 In similar triangles PERIMETERS proportional to sides:
AB CA BC + B C K AB KL CA MK BC LM KL MK LM + M KL AB MK CA LM BC + OR L STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19 Given: JT=20, JU=30, TU=70, AND JK=90, what is the perimeter of IJK
Given: JT=20, JU=30, TU=70, AND JK=90, what is the perimeter of IJK? Suppose TJU IJK. S R J U T I K J I K S U T R 20 30 90 70 JK PERIMETER IJK = JU JT+JU+TU 90 30 PERIMETER IJK = 120 PERIMETER IJK = 3 (120) (120) PERIMETER IJK = 360 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20 A AB AC B C BD DC D STANDARDS 4 and 5
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21 JS is an angle bisector. JI= X+5, JK=X+3, IS=3, and SK=2
JS is an angle bisector. JI= X+5, JK=X+3, IS=3, and SK=2. Find the value for X. Suppose IJK TJU. S R J U T I K X+5 X+3 IS JI X+5 X+3 2 3 = 3 2 SK JK 3(X+3) = 2(X+5) 3X + 9 = 2X + 10 3X = 2X + 1 -2X -2X X = 1 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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