1. Concurrencies for Medians, Altitudes, and Bisectors
Part 2
Concurrencies for Medians, Altitudes, and Bisectors

2. Theorem 5.1 Perpendicular Bisector Theorem
If a point is on the perpendicular bisector of a segment, then it’s equidistant from the endpoints of the segment. C
If segment CP is the perpendicular bisector of segment AB, then segment CA = segment CB. B A P

3. Theorem 5.2 Converse of the Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it’s on the perpendicular bisector of the segment. D
If segment DA = Segment DB, then D lies on the perpendicular bisector of segment AB. B A P

4. Ex 1: In the diagram shown, line MN is the perpendicular bisector of segment ST.Q T M N 12
a. What segment lengths in the diagram are equal?
b. What Theorem explains why Q is on line MN?

6. Check HW from Part 1

7. Median of a triangle
Segment AD is a median of  ABC
Median of a Triangle- segment whose endpoints are a vertex of a triangle and the midpoint of the opposite side.
The three medians of a triangle are ALWAYS concurrent!

8. When three or more lines (or rays or segments) intersect at a common point, then they are called concurrent lines (or rays or segments). The point of intersection of the lines is called the point of concurrency.

9. Centroid
Centroid- the point of concurrency of the 3 medians of a triangle (in this example point X)
It’s always inside the triangle
It’s also the balancing point of the triangle.

10. Theorem 5.7 - Centroid Theorem
The medians of a triangle intersect at a pt that is 2/3 of the distance from each vertex to the midpoint of the opposite side.
If P is the centroid of
ABC, then
AP = 2/3 AD,
BP = 2/3 BF, &
CP = 2/3 CE

11. Ex 1: P is the centroid of QRS shown below and PT = 5. Find RT and RP.
Because P is the
centroid, RP = 2/3RT.
PT = RT-RP = 1/3RT
Substitute 5 for PT,
5 = 1/3RT, so RT = 15
RP = 2/3RT = 2/3(15) = 10
So, RP = 10 and RT = 15

14. Method 1: Find the coordinates of the centroid of JKL with vertices L(3, 6) K(5, 2) J(7, 10). (Hint: graph it 1st!) ( ) ( )
=N(5,8)
& Distance of NK = 6
PK = (2/3)6 = 4
So, P(5,6)

15. Method 2: Find the coordinates of the centroid of JKL with vertices L(3, 6) K(5, 2) J(7, 10).
Finding the averages of
your coordinate points

17. Applications

18. Median and Centroid WS

19. More Points of Concurrency

20. Perpendicular Bisectors and Angle Bisectors
Circumscribed – a circle drawn
about another shape.
Inscribed – a circle drawn within
the boundaries of another shape.

21. Perpendicular Bisectors Theorem 5.3 – Circumcenter Theorem
The perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.
PA = PB = PC

22. Concurrency - Right Triangle
The point of concurrency is on the triangle.
Concurrency - Acute Triangle
The point of concurrency is inside the triangle.
Concurrency - Obtuse Triangle
The point of concurrency is outside the triangle.

23. Angle Bisector
A ray, segment, or line that breaks an angle into two adjacent angles that are congruent. ) )

24. Theorem 5.4 Angle Bisector Theorem
If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle. (must be perpendicular!)
If m< BAD = m< CAD, then DB = DC. B D ) A )
DB = DC C

25. Theorem 5.4 Converse of the Angle Bisector Theorem
If a point is in the interior of an angle and is on equidistant from the two sides of the angle, then it lies on the bisector of the angle.
If DB = DC, then m< BAD = m< CAD (Same picture from previous theorem)

26. Ex 2: Ray PM is the angle bisector of <LPN
a) What is the relationship between <LPM and <NPM?
b) If MN=17, what is ML?

27. Theorem 5.6 – Incenter Theorem
The angle bisectors of a triangle intersect at a point called the incenter that is equidistant from the sides of the triangle.
PD = PE = PF

28. Ex 2: Using Angle Bisectors
The angle bisectors of ∆MNP meet at point L.
What segments are congruent?
Find LQ and LR.

29. Where’s the Point of Concurrency for Angle Bisectors (Incenter)?
Acute Triangle: interior
Right Triangle: interior
Obtuse Triangle: interior

30. Altitude of a triangle- the  seg form a vertex to the opposite side of the line that contains the opposite side. Orthocenter- the lines containing the three altitudes of a triangle are concurrent, pt of concurrency is the orthocenter
Orthocenters:
Acute: Interior Obtuse: Exterior Right: On

31. Orthocenter Theorem
The lines containing the altitudes of a triangle are concurrent. This point orf concurrency is called the orthocenter.
If segments AE, BF, and CD are the altitudes of ABC, then the lines AE, BF, and CD intersect at some point H.

36. Calculating Points of Concurrency Packet