-
Symbol Defintion
Sqr(x) = Square root of x
GE 15 01. What is in-center
Definition
- Bisectors of angles of triangle are concurrent at a point I
- The point is called in-center
- Three lines meet at one point is called concurrent
- A line biscets an angle equally which is bisector of an angle
- Bisector theory of angle
- point on bisector of an angle has same distance from the two rays
- If point between the rays of angle has same distance, point is on bisector
Go to Begin
GE 15 02. bisectors of angles of triangle are concurrent
Construction
- Draw triangle ABC
- Let bisector of angle A and bisector of angle B meet at I
- Join C and I as a line
- We want to prove that IC is bisector of angle C
- Draw IF perpendicular to AB, ID perpendicular to BC
- IA = IB and IB = IC (bisector theory)
- Hence IA = IC
- Hence IC is bisector of angle C (bisector theory)
- Hence bisectors of angles of triangle are concurrent
Go to Begin
GE 15 03. In-circle tangents the sides of triangle
Construction
- In Q02, ID = IE = IF = in-radius = r
- Hence use I as center and r as radius to make a circle
- The circle will tangent the sides of triangle
- Let s = (a + b + c)/2
- AE = AF, BD = BF and CD = CE
- AB = c = AF + BF, BC = a =BD + CD and CA = b = CE + AE
- Hence a + b + c = 2*AF + 2*(BD + CD)
- Hence AF = (a + B + c)/2 - a = (s - a)
Go to Begin
GE 15 04. Locus of in-center
Conditions
- Let A and B are fixed and C is moving with angle ACB = constant
- Find locus of in-center
-
Diagram of geometry
Program 10 02
- We can proof that angle AIB = pi - A/2 - B/2
- Angle A + angle B = pi - C, Hence AIB = pi/2 + C/2
- Since angle AIB is constant and point A and point B are fixed
- Hence point I will move along arc AIB (circum-circle of tringle AIB)
Go to Begin
GE 15 05. Coordinate geometry : bisector of angles of triangle are concurrent
Reference
-
Coordinate geometry
AN 11 03
Go to Begin
GE 15 06. Find coordinate of in-center
Use tangent theory
- We can prove that AF = AE = s - a = length of tangent from A to F or A to E
- Where s = (a+b+c)/2 and a,b,c are lenght of sides
- Method 1
- Let AB be in horizontal direction and A be the origin
- The center I(xi,yi)
- xi = (s - a)
- yi = r*tan(A/2)
- Method 2
- Find equations of two bisectors using slope and points
- Find intersections of two equations
- The intersection is the in-center
Go to Begin
GE 15 07. What is ex-center ?
Definition
- Bisectors of one internal angle A and external angles B and C
- They are concurrent at a point I
- The point is called ex-center
- A triangle has three ex-centers
- Three lines meet at one point is called concurrent
- A line biscets an angle equally which is bisector of an angle
- Bisector theory of angle
- point on bisector of an angle has same distance from the two rays
- If point between the rays of angle has same distance, point is on bisector
Go to Begin
GE 15 08. Bisectors of angles of triangle are concurrent
Construction
- Draw triangle ABC
- Let bisector of external angles B and C meet at I
- Join A and I as a line
- We want to prove that AI is bisector of angle A
- Draw IF perpendicular to AB, ID perpendicular to BC
- IF = ID (bisector theory)
- Draw ID perpendicular to BC, IE perpendicular to CA
- IE = ID (bisector theory)
- Hence IE = IF
- Hence AI is bisector of angle A (bisector theory)
- Hence bisectors of angles of triangle are concurrent
Go to Begin
GE 15 09. In-circle tangents the sides of triangle
Construction
- In Q02, ID = IE = IF = in-radius = r1
- Hence use I as center and r1 as radius to make a circle
- The circle will tangent the sides of triangle
- Let s = (a + b + c)/2
- AF = AD, BF = BD and CD = CE
- AF + AE = AB + BF + AC + CE
- 2*AF = AB + AC + (BF + CE)
- 2*AF = a + b + c
- AF = (a + b + c)/2
Go to Begin
GE 15 10. Locus of ex-center
Conditions
- Let A and B are fixed and C is moving with angle ACB = constant
- Find locus of ex-center
-
Diagram of geometry
Program 10 02
- We can proof that
- Angle AIB = pi/2 - A/2 - B/2
- Angle AIB = pi/2 - C/2
- Since angle AIB is constant and point A and point B are fixed
- Hence point I will move along arc AIB (circum-circle of tringle AIB)
Go to Begin
GE 15 11. Coordinate geometry : bisectors of angles of triangle concurrent
Reference
-
Coordinate geometry
AN 11 04
Go to Begin
GE 15 12. Find coordinate of ex-center
Use tangent theory
- We can prove that AF = AE = s = length of tangent from A to F or A to E
- Where s = (a+b+c)/2 and a,b,c are lenght of sides
- Method 1
- Let AB be in horizontal direction and A be the origin
- The center I(xi,yi)
- xi = s where s = (a + b + c)/2
- yi = r1*tan(A/2) = s*tan(A/2)
- Method 2
- Find equations of two bisectors using slope and point
- Find intersections of two equations
- The intersection is the ex-center
Go to Begin
GE 15 00. Outlines
In-center
- Defintion : Three angle bisectors of triangle are concurrent
- In-circle : It is a circle with 3 sides of triangle as tangent
- Tnagent from vertex to in-circle
- AF = AE = (s - a)
- BD = CF = (s - b)
- CE = BD = (s - c)
- AF = AE = s
- BD = CF = s
- CE = BD = s
Go to Begin