Mathematics Dictionary
Dr. K. G. Shih
In-center
Subjects
Symbol Defintion
Sqr(x) = Square root of x
Q01 |
- What is in-center of a triangle ?
Q02 |
- Prove that bisectors of angles of triangle are concurrent
Q03 |
- In-circle tangents the sides of triangle
Q04 |
- Locus of in-center
Q05 |
- Using coordinate geometry Prove that bisector of angles of triangle concurrent
Q06 |
- Coordinates of in-center
Q07 |
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Q08 |
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Q09 |
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Q10 |
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Answers
Q01. What is in-center
Definition
Bisectors of angles of triangle are concurrent at a point I
The point is called in-center
What is concurrent ?
Three lines meet at one point is called concurrent
What is bisctor of an angle ?
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
diagram
Diagram of geometry
Program 03 02
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Q02. 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
Proof
Draw IF perpendicular to AB, ID perpendicular to BC and IE perpendicuar to CA
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
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Q03. In-circle tangents the sides of triangle
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
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Q04. Locus of circum-center
Conditions
Let A and B are fixed and C is moving with angle ACB = constant
Find locus of in-center
Diagram
Diagram of geometry
Program 10 02
Proof
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
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Q05. Using coordinate geometry Prove bisector of angles of triangle concurrent
Reference
Coordinate geometry
Program 11 ??
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Q06. Find coordinate of in-center
Use tangent theory
We can prove that AF = AE = s - a if the length of tangent from A to F
Where s = (a+b+c)/2 and a,b,c are lenght of sides
Find coordinate of in-center
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 point
Find intersections of two equations
The intersection is the circum-center
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Q07. Answer
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Q08. Answer
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Q09. Answer
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Q10. Answer
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