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Lesson 5-1 Bisectors, Medians and Altitudes. Objectives Identify and use perpendicular bisectors and angle bisectors in triangles Identify and use medians.

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Presentation on theme: "Lesson 5-1 Bisectors, Medians and Altitudes. Objectives Identify and use perpendicular bisectors and angle bisectors in triangles Identify and use medians."— Presentation transcript:

1 Lesson 5-1 Bisectors, Medians and Altitudes

2 Objectives Identify and use perpendicular bisectors and angle bisectors in triangles Identify and use medians and altitudes in triangles

3 Vocabulary Concurrent lines – three or more lines that intersect at a common point Point of concurrency – the intersection point of three or more lines Perpendicular bisector – passes through the midpoint of the segment (triangle side) and is perpendicular to the segment Median – segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex Altitude – a segment from a vertex to the line containing the opposite side and perpendicular to the line containing that side

4 Vocabulary Circumcenter – the point of concurrency of the perpendicular bisectors of a triangle; the center of the largest circle that contains the triangle’s vertices Centroid – the point of concurrency for the medians of a triangle; point of balance for any triangle Incenter – the point of concurrency for the angle bisectors of a triangle; center of the largest circle that can be drawn inside the triangle Orthocenter – intersection point of the altitudes of a triangle; no special significance

5 Theorems Theorem 5.1 – Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. Theorem 5.2 – Any point equidistant from the endpoints of the segments lies on the perpendicular bisector of a segment. Theorem 5.3, Circumcenter Theorem – The circumcenter of a triangle is equidistant from the vertices of the triangle. Theorem 5.4 – Any point on the angle bisector is equidistant from the sides of the triangle. Theorem 5.5 – Any point equidistant from the sides of an angle lies on the angle bisector. Theorem 5.6, Incenter Theorem – The incenter of a triangle is equidistant from each side of the triangle. Theorem 5.7, Centroid Theorem – The centroid of a triangle is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median.

6 Triangles – Perpendicular Bisectors C Circumcenter Note: from Circumcenter Theorem: AP = BP = CP Midpoint of AB Midpoint of BC Midpoint of AC Z Y X P B A Circumcenter is equidistant from the vertices

7 Triangles – Angle Bisectors B C A Incenter Note: from Incenter Theorem: QX = QY = QZ Z X Y Q Incenter is equidistant from the sides

8 Triangles – Medians C Midpoint of AB Midpoint of BC Midpoint of AC Centroid Median from B M Z Note: from Centroid theorem BM = 2/3 BZ Y X Centroid is the point of balance in any triangle B A

9 Triangles – Altitudes C X Z Y Orthocenter Altitude from B Orthocenter has no special significance for us B A Note: Altitude is the shortest distance from a vertex to the line opposite it

10 Special Segments in Triangles NameType Point of Concurrency Center Special Quality From / To Perpendicular bisector Line, segment or ray Circumcenter Equidistant from vertices None midpoint of segment Angle bisector Line, segment or ray Incenter Equidistant from sides Vertex none MediansegmentCentroid Center of Gravity Vertex midpoint of segment AltitudesegmentOrthocenternoneVertex none

11 Location of Point of Concurrency NamePoint of Concurrency Triangle Classification AcuteRightObtuse Perpendicular bisectorCircumcenterInsidehypotenuseOutside Angle bisectorIncenterInside MedianCentroidInside AltitudeOrthocenterInsideVertex - 90Outside

12 Given: Find: 1. 2. 3. 4. 5. 6. 7. 8. m  DGE

13 Find: Given:. m  ADC EXAMPLE 2

14 ALGEBRA Points U, V, and W are the midpoints of respectively. Find a, b, and c. Find a. Segment Addition Postulate Centroid Theorem Substitution Multiply each side by 3 and simplify. Subtract 14.8 from each side. Divide each side by 4. EXAMPLE 3

15 Find b. Segment Addition Postulate Centroid Theorem Substitution Multiply each side by 3 and simplify. Subtract 6b from each side. Divide each side by 3. Subtract 6 from each side. CONT.

16 Find c. Segment Addition Postulate Centroid Theorem Substitution Multiply each side by 3 and simplify. Subtract 30.4 from each side. Divide each side by 10. Answer: CONT.

17 ALGEBRA Points T, H, and G are the midpoints of respectively. Find w, x, and y. Answer: EXAMPLE 4

18 Summary & Homework Summary: –Perpendicular bisectors, angle bisectors, medians and altitudes of a triangle are all special segments in triangles –Perpendiculars and altitudes form right angles –Perpendiculars and medians go to midpoints –Angle bisector cuts angle in half Homework: page 242 (6, 13-16, 21-26)


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