1. 1. For A(–4, 8) and B(5, 8), find the midpoint of AB.
ANSWER 1 2
, 8
2. For A(–3, 2) and B(4, –1), find the length of AB.
ANSWER 58
3. For A(0, 4) and C(18, 4), find the length of AB, where B is a point the distance from A to C. 2 3
ANSWER 12

2. Use properties of special segments in triangles.
Target
Use properties of special segments in triangles.
You will…
Use medians and altitudes of triangles.

3. Vocabulary
median of a triangle – a segment from a vertex to a midpoint of the opposite side.
Concurrency of Medians of a Triangle Theorem 5.8 – The medians of a triangle intersect at a point (centroid) that is 2/3 the distance from each vertex to the midpoint of the opposite side.
The lengths of the 3 segments of each median are in a 3:2:1 ratio.
Median AE
AE:AP:PE = 3:2:1 = 3x:2x:1x
Median BF
BF:BP:PF = 3:2:1
Median CD
CD:CP:PD = 3:2:1

4. Vocabulary
altitude of a triangle – a segment from a vertex perpendicular to the line containing the opposite side.
Concurrency of Altitudes of a Triangle Theorem 5.9 –
The lines containing the altitudes of a triangle are concurrent in a point (orthocenter).

5. located by the perpendicular bisectors of the sides
Review
Circumcenter
located by the perpendicular bisectors of the sides
equidistant from the vertices of the triangle
center of the circumscribed circle
Incenter
located by the angle bisectors of the triangle
equidistant from the sides of the triangle
center of the inscribed circle
Centroid
located by the medians of the triangle
divides the median (vertex to centroid) in a 2:1 ratio
balance point
Orthocenter
located by the lines containing the altitudes of the triangle

6. Use the centroid of a triangle
EXAMPLE 1
Use the centroid of a triangle
In RST, Q is the centroid and SQ = 8. Find QW and SW.
SOLUTION
SQ = 2 3 SW
Concurrency of Medians of a Triangle Theorem
8 = 2 3 SW
Substitute 8 for SQ.
Multiply each side by the reciprocal, . 3 2
12 = SW
Then QW =
SW – SQ =
12 – 8 = 4.
So, QW = 4 and SW = 12.

7. EXAMPLE 2
Standardized Test Practice
SOLUTION
Sketch FGH. Then use the Midpoint Formula to find the midpoint K of FH and sketch median GK .
K( ) =
2 + 6 , 2
K(4, 3)
The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side.

8. EXAMPLE 2
Standardized Test Practice
The distance from vertex G(4, 9) to K(4, 3) is
9 – 3 = 6 units. So, the centroid is (6) = 4 units
down from G on GK . 2 3
The coordinates of the centroid P are (4, 9 – 4), or (4, 5).
The correct answer is B.
Median GK
GK:GP:PK = 3:2:1 = 3x:2x:1x
6:GP:PK = 3x:2x:1x
6 = 3x
2 = x
GP = 2x = 4; PK = 1x = 2

9. GUIDED PRACTICE
for Examples 1 and 2
There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P. 1.
If SC = 2100 feet, find PS and PC.
700 ft, 1400 ft
ANSWER

10. GUIDED PRACTICE
for Examples 1 and 2
There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P. 2.
If BT = 1000 feet, find TC and BC.
1000 ft, 2000 ft
ANSWER

11. GUIDED PRACTICE
for Examples 1 and 2
There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P. 3.
If PT = 800 feet, find PA and TA.
1600 ft, 2400 ft
ANSWER

12. EXAMPLE 3
Find the orthocenter
Find the orthocenter P in an acute, a right, and an obtuse triangle.
SOLUTION
Right triangle
P is on triangle.
Acute triangle
P is inside triangle.
Obtuse triangle
P is outside triangle.

13. EXAMPLE 4
Prove a property of isosceles triangles
Prove that the median to the base of an isosceles triangle is an altitude.
GIVEN :
ABC is isosceles, with base AC . BD is the median to base AC .
PROVE :
BD is an altitude of ABC.
Proof : Legs AB and BC of isosceles ABC are congruent.
CD AD because BD is the median to AC . Also, BD BD . Therefore, ABD CBD by the SSS Congruence Postulate.
ADB  CDB because corresponding parts of  s are . Also, ADB and CDB are a linear pair. BD and AC intersect to form a linear pair of congruent angles, so BD  AC and BD is an altitude of ABC.

14. GUIDED PRACTICE
for Examples 3 and 4 5.
WHAT IF?
In Example 4, suppose you wanted to show that median BD is also an angle bisector. How would your proof be different?
CBD
ABD
By SSS making
which leads to BD being an angle bisector.
ANSWER

15. GUIDED PRACTICE
for Examples 3 and 4 4.
Copy the triangle in Example 4 and find its orthocenter.
SOLUTION

16. GUIDED PRACTICE
for Examples 3 and 4 6.
Triangle PQR is an isosceles triangle and segment OQ is an altitude. What else do you know about OQ ? What are the coordinates of P?
OQ is also a perpendicular bisector, angle bisector, and median; (-h, 0).
ANSWER