Figure 127 : Circum-center
Q01. Diagram
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Q02. Circum-center
Construction
- Draw a triangle ABC
- Draw bisector of side AB
- Draw bisector of side BC
- Draw bisector of side CA
- Bisectors are concurrent at P which is the circum-center
- Since EH is bisector of AB, hence EA = EB
- Since EF is bisector of BC, hence EB = EC
- Since EA = EC, hence ER is bisector of AC
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Q3. Sine law
- Let R be the radius of circum-circle
- Then a = 2*R*sin(A)
- Then b = 2*R*sin(B)
- Then c = 2*R*sin(C)
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Q04. Reference
- Program 02 05
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Program 03 02
- See keyword : Circum-center
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Q5. Questions
- 1. What is circum-center ?
- 2. Solve triangle if a = 5, b = 4 and A = 30 degrees
- 3. Draw triangle ABC and circum-center E. Find radius of circum-circle
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