1. EOC Review #5 Thursday Given: Endpoint (2, 3); Midpoint (3, -4)
Find the missing endpoint.
EOC REVIEW QUIZ TOMORROW 

2. Honors H.W. #23 pg. 246
#2-32 and 38 (even)

3. H.W. QUESTIONS?

4. 38.) In GHJ, K(2,3) is the midpoint of GH, L(4,1) is the midpoint of HJ, and M(6,2) is the midpoint of GJ.

5. Bisectors in Triangles Toolkit #5.2
Today’s Goal:
To use properties of perpendicular bisectors and angle bisectors.

6. Perpendicular Bisector: equidistant from the endpoints of a segment (cuts SEGMENT in two EQUAL parts), and makes a RIGHT ANGLE with that segment.
When three perpendicular bisectors meet, that POINT is called the CIRCUMCENTER.

7. Find the set of points on the map of D. C
Find the set of points on the map of D.C. that are equidistant from the Jefferson Memorial and the White House.

8. Ex.1: CD is the perpendicular bisector of AB. Find CA and DB.

9. Angle Bisector: equidistant from the sides of a triangle – cuts an ANGLE into TWO EQUAL parts.
When three angle bisectors meet, the POINT is called the INCENTER.

10. Ex.2(a): Find the value of x, then find FD and FB.

11. Ex.2(b): Check Understanding
According to the diagram, how far is K from EH? From ED?
What can you conclude about EK?
Find the value of x.
Find mDEH.

12. In-Class Practice (Part I) Geometry Book
Pg. 251
#’s 1-4, 6-11

13. Solutions – Part I AC is the  bis. of BD 15 18 8 x = 12, JK = JM = 17
y = 3, ST = TU = 15
HL is the  bis. of JHG.
y = 9, mFHL=54, mKHL=54 27
Point E is on the bisector of KHF.

14. In-Class Practice (Part II) Geometry Book
Pg. 252
#’s 12-15, 18-25

15. Solutions – Part II 12 4 16 5 10 7 14 5 10
Isosceles

16. In-Class Practice (Part III) Geometry Book
Pg. 252
#’s 28-30
(Draw picture!)

17. Solutions – Part III No, A is not equidistant from the sides of X
Yes, AX bisects TXR
Yes, A is equidistant from the side of X.

18. pg. 253 #40

19. Bisectors & Graphing Ch. 5.2 Extension
Today’s Goal(s)
To investigate perpendicular and angle bisectors on the coordinate plane.

20. Example: Given points A(1,3), B(5,1), and C(4,4), does C lie on the  bisector of AB?
Plot points first!
Then determine whether AC = BC.

21. #5 Perpendicular Bisector a. (-2,7) b. (-1,6) c. (0,5)

22. #6 Angle Bisector a. (6,5) b. (7,8) c. (4,4)

23. You are given the points A(4, 8), O(0,0), and B(12, 0).

24. What did the waiter say when he delivered the potato?

25. What was Humpty-Dumpty’s Cause of death?

26. Concurrent Lines, Medians, and Altitudes Toolkit 5.3
Today’s Goal:
To identify properties of medians and altitudes of a triangle.

27. STOP and THINK! What does concurrent mean?
When three or more lines intersect in one point, the point at which they intersect is the point of concurrency.

28. Median of a triangle: a segment that connects a VERTEX to the MIDPOINT of the opposite side.
When three medians meet, the POINT is called the CENTROID.

29. The medians of a triangle are concurrent at a point where the distance from the point to the vertex is TWICE the distance from the point to the side.

30. Ex.1(a): Finding Lengths of Medians
D is the centroid of ABC and DE = 6.
Find BE.

31. Ex.1(b): Finding Lengths of Medians
You Try…
M is the centroid of WDR, and WM = 16.
Find WX.

32. Altitude of a triangle: a segment that is PERPENDICULAR to a side, and connects a VERTEX to the opposite side.
When three altitudes meet, the POINT is called the ORTHOCENTER.

33. 4 Points of Concurrency Summary Helpful Hint!
CM PB
IN AB
C M
O A

34. Other Uses! Perpendicular Bisectors
Pt. of Concurrency:
The circle is circumscribed
about ABC.

35. Other Uses! Angle Bisectors
Pt. of Concurrency:
The circle is inscribed
in ABC.

36. Other Uses! Medians Pt. of Concurrency: “Balance Point”
Also called the center of gravity of a triangle because it is the point where a triangular shape will balance.

37. Other Uses! Altitudes
Pt. of Concurrency:
“Heights” of a triangle.

38. Ex.2(a.): Find the center of the circle that circumscribes XYZ.

39. Ex.2(b.): Find the center of the circle that circumscribes XYZ.
You Try…
Ex.2(b.): Find the center of the circle that circumscribes XYZ.
X(0,0), Y(0,6), Z(4,0)