5.1 Notes
Orthocenter & Altitudes
I don’t tell you much other than an angles like AHE  90
Degrees. So in that picture if a problem said AHE  3 X  5
you would set up 3X+5 = 90
You can also set up A2  B2  C 2 so AE 2  EH 2  AH 2
Orthocenter  Intersection of 3 Altitudes
I don’t tell you much other than an
  90 so if an angle is labeled X+5
you would set up X+5 = 90
You can also set up A2  B2  C 2
Altitudes  Creates 2 Right Angles
Circumcenter & Perpendicular Bisectors
Perpendicular Part Means I
can draw Have Right angles
and Right Triangles like
AED  BED  90 &
Right AED & Right BED and
can use A2  B2  C 2
The Bisector Part Means each side of the
Triangle is Cut in Half.
The Circle Outside means the Distance from
the Circumcenter to the Triangle Vertex are
always he same and are Radius to the Circle
Circumcenter  Intersection of 3 Perpendicular Bisectors
Circumcenter  You can draw a Circle OUTSIDE the triangle
Notice the Perpendicular Bisectors
Are DF DE and DG these do not draw
A line back to Triangle Vertex
Notice we can Draw New Dotted
Lines DB DA DC in and they are
Radius of the Circle
I tell you a TON. I tell you about
where to draw a right Angle
I tell you a TON. I tell you the side is cut in half. t
AED  90 so if I tell you
So if AE=EB=1/2AB and I tell you AE=10 & EB=X+3
then X+3=10
AED  4 X  10 then you set
up 4 X  10  90
So if AE = 10 & AB = 2X+5 then 10*2 = 2X+5
AE=EB=1/2AB and I tell
AE=10
Perpendicular
youCreates
2 Right Angles
& EB=X+3 then X+3=10
Bisector
Half Side AX = Half Side XB
D
E
You can basically set up multiple kinds of problems
Side Problems using Bisector AE = EB or AE = ½ AB
Side Problems using Pythogream A2  B2  C 2 BE 2  ED2  BD2
Angle Problems where you Solve the Right Angle
Angle Problems where you solve the
AED  90 and AED  9 X  10
EDB  40 EBD  40 EDB  2 X  so 40+90+2X=180
Incenter & Angle Bisectors
Side Problems
Angle Problems
Distance from Incenter to a Side is always
same (Radius )
1
IAF  IAB  ABC So
2
ABC  40 & IAB  2 X
if IE = 7 and IF = 4X+3 then 7=4X+3
Then
1
(40)  2 X
2
Incenter  3 Angle Bisectors
 Circle Inside Triangle
I tell you the two half angles are equal.
So if EB bisects <ABC then <ABE=<CBE=1/2<ABC
So if <ABC = 70 then two angles are 35 and 35 OR
If ABE = 40 and CBE = 3X+2 then 3X+2=40 and <ABC =80
A
Or
Angle Bisector – Cuts an Angle in Half
C
B
E
Centroids and Medians
Centroid’s use the Small Piece is “1/3” Big Piece is “2/3”
So if I said SV = 10 then VR = 20 and SR =30
If I said VT = 5X and VP =40. VP is big piece so you know small piece
is 20. So 5X=20
Centroid  3 Medians
I tell you the two half SIDES are equal.
So if if AD is Median in Triangle ABC then BD=DC=1/2 BC
So if BC = 10 then two half sides are 5 and 5 OR
If BD = 10 and CBE = 5X+5 then 5X+5=10 and BC=20
A
Or
Median – Cuts a side in half
B
D
C
Really there are No easy Answers to this stuff.
You need to Know the Terms
You need to Know What to Do with “Side” Problems & Angle
Problems
Side Problems Might Be
1) Centroid 1/3 (Small) 2/3 (Big) Whole Median
2) Median/Perpendicular Bisector where a side of triangle is Cut
in Half so BD=DC=1/2BC
3) Circumcenter Problem where Circumcenter to Triangle Vertex
(points) are all equal (Radius)
4) Incenter Problem where Incenter to Triangle Side are all
equal (Radius)
5) Pythogrean Theorem Problem where you use A  B  C to find
the Distance. However becareful for a Circumcenter that’s
“C” for Incenter that’s “B”
2
2
2
If you have Coordinate Points on a Graph like (4,4) you can
6) You can use Distance Formula
7) You can use Midpoint Formula to find the Midpoints to draw
the Medians
Angle Problems
1) You can use Angle Bisectors to set up
1
IAF  IAB  ABC type
2
problems
2) You can use Altitudes or Perpendicular Bisectors to draw in a Right Angle and if they give you a
Equation for that Right Angle you would just set them equal AXB  90 and AXB  4 X  10
then 4X+10 = 90
3) You can use Altitudes or Perpendicular Bisectors to Make Right Triangles which means you Add the
3 angles together (including the 1 that is 90) and set equal to 180