Figure 128 : In-center
Q01. Diagram
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Q02. In-center
Construction
- Draw a triangle ABC
- draw bisector of angle A
- Draw bisector of angle B
- Draw bisector of angle C
- Bisectors are concurrent and meet at I which is the in-center
- Let IP perpendicular to AB
- Let IQ perpendicular to AC
- Let IR perpendicular to BC
- Since AI is angle bisector, hence IP = IQ
- Since BI is angle bisector, hence IP = IR
- Since IR = IQ, Hence CI is also angle bisector
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Q03. Locus of in-center
Question
- A and B are fixed
- Angle ACB keeps constant if C moves
- Find locus of in-center I
- Triangle ACP and BCP
- Angle AIP = (angle A)/2 + (angle C)/2
- Angle BIP = (angle B)/2 + (angle C)/2
- Hence Angle AIB = angle AIP + angle BIP = (A + B + 2*C)/2
- Since Angle A + B + C = pi
- Hence Angle AIB = (pi + C)/2 = constant
- Point A and B are constant.
- Hence locus of I is arc AIB on circle
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Q04. Tangent from A to in-circle
Thangent length from A to circle AP = (s - a)
- Proof : See keyword in-circle
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Q05. Reference
- Program 02 06
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Program 03 02
- See keyword in-center
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Q06. Questions
- 1. What is in-circle ?
- 2. Find radius of in-circle
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