Download presentation

Presentation is loading. Please wait.

1
**5.1 Bisector, Medians, and Altitudes**

4
B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment. Example 1: Find x, if AD is a perpendicular bisector. 6x - 5 3x + 16 A B C D

5
C. Theorem 5.3: Circumcenter Theorem The circumcenter of a triangle is equidistant from the vertices of the triangle. Example 2 Find x and y, if BE = 24, AE = 3x – 6 and CE = 5y + 4 A B C E

6
D. Theorem 5.4: If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle. Example 3 Find x, if AC is an angle bisector of DAB and DC = x + 4 and BC = 2x – 5. D C A B

7
E. Theorem 5.7: Centroid Theorem The distance from the vertex to the point of concurrency is equal to 2 times the distance from the point of concurrency to the midpoint of the opposite side. Example 4 Point A is a centroid of ∆DEF. Find x, y, and z. D S E T F U A SA = 4z EA = y TA = 2x – 5 FA = 4.6 UA = 2.9 DA = 6

Similar presentations

OK

5.3 Concurrent Lines, Medians, and Altitudes Stand 0_23456789 Can you figure out the puzzle below??? No one understands!

5.3 Concurrent Lines, Medians, and Altitudes Stand 0_23456789 Can you figure out the puzzle below??? No one understands!

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on indian defence forces Ppt on percentage for class 6 Download ppt on wifi technology Ppt on nitrogen cycle and nitrogen fixation reaction Ppt on channels of distribution for a product Ppt on satellite orbiting Ppt on mahatma gandhi quotes Free ppt on personality development Ppt on relations and functions for class 11th sample Free ppt on forest society and colonialism in india