# Sangaku with Angle between a Tangent and a Chord

The applet below illustrates a 1896 Sangaku from the Ibaragi prefecture ([Fukagawa & Pedoe], problem 1.1.4, the tablet has vanished.)

The tangents at points A and B on a given circle Γ intersect at C. Show that the incenter I of triangle ABC lies on Γ.

What if applet does not run? |

### References

- H. Fukagawa, D. Pedoe,
*Japanese Temple Geometry Problems*, The Charles Babbage Research Center, Winnipeg, 1989Write to:

Charles Babbage Research Center

P.O. Box 272, St. Norbert Postal Station

Winnipeg, MB

Canada R3V 1L6

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Copyright © 1996-2018 Alexander Bogomolny

The tangents at points A and B on a given circle Γ intersect at C. Show that the incenter I of triangle ABC lies on Γ

Let I' be the midpoint of the small arc AB of Γ.

The arcs AI' and BI' are equal. The inscribed angle ABI' equals half the arc AI', angle BAI' equals half the arc BI'. The two angles CAI' and CBI' between the tangents to Γ and the chords AI' and BI' equal respectively half the arcs AI' and BI'. It follows that

## Sangaku

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- Bottema Shatters Japan's Seclusion
- Chain of Circles on a Chord
- Circles and Semicircles in Rectangle
- Circles in a Circular Segment
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- Equal Incircles Theorem
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- Sangaku with Quadratic Optimization
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- Sangakus with a Mixtilinear Circle
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- Two Circles in an Angle
- Two Sangaku with Equal Incircles
- Another Sangaku in Square
- Sangaku via Peru
- FJG Capitan's Sangaku

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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