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Symbol Defintion
Sqr(x) = Square root of x
Q01. What is circum-center
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- Bisectors of sides of triangle are concurrent at a point I
- The point is called circum-center
- Three lines meet at one point is called concurrent
- A line biscets a line equally which is bisector of line
- Bisector theory of line
- point on bisector of a line has same distance from the ends of line
- If point has same distance from ends of line, the point is on bisector
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Diagram of geometry
Program 03 02
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Q02. bisectors of sides of triangle are concurrent
Construction
- Draw triangle ABC
- Let D be mid point of BC, E be midpoint of CA and F be mid point of AB
- Draw draw bisector of AB and bisector of BC intersecting at I
- We want to prove that bisector of CA will pass point of I
- Since ID is bisector of BC, hence IB = IC (bisector theory)
- Since IF is bisector of AB, hence IA = IB (bisector theory)
- Hence IA = IC
- Hence IE is bisector of CA (bisector theory)
- Hence bisector of CA also passes point I
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Q03. Circum-circle passes the vertices of triangle
- In Q02, IA = IB = IC = circum-radius = R
- Hence use I as center and R as radius to make a circle
- The circle will pass point A, B, C
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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 circum-center
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Diagram of geometry
Program 10 04
- IA = IB = IC and AB is fixed
- A, B, C are always on same circle
- Hence I will not move
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Q05. Using coordinate geometry Prove that bisector of sides of triangle concurrent
Reference
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Coordinate geometry
Program 11 ??
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Q06. Sine law and circum-circle
Sine law
- a/sin(A) = b/sin(B) = c/sin(C) = 2*R
- Method 1
- Let AB be in horizontal direction and A be the origin
- The center I(xi,yi)
- xi = AB/2 and y = a/(2*sin(A))
- 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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