Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Tuesday, December 10, 2019

### Dynamic Geometry 1452: Japanese Theorem, Sangaku

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https://photos.app.goo.gl/QpkuTkRDCyewn31j8

ReplyDeleteLet I and J are the midpoints of arc AD and arc CD

Define points P and Q as shown on the sketch

quadrilateral AA1D1D is cyclic with I is the circumcenter.

quadrilateral DD1C1C is cyclic with J is the circumcenter.

Angle(AD,BC) = ½(arc DC- Arc BA)= ½( ∠ (COD)- ∠ (AOB))

Angle (AD, A1D1)=1/2( arc DD1- arc AA1)= ½(∠ (DID1)- ∠ (AIA1))

Since ∠ (DID1)=1/2 ∠ (COD)=> Angle (AD, A1D1)=1/2 Angle(AD,BC)

So A1D1 is parallel to the angle bisector of angle(BC, AD)=>projections of A1D1 to BC and AD will have equal lengths

1. Per the result of problem 359 we have A1D1 ⊥D1C1

Similarly, for other corners => A1B1C1D1 is a rectangle

2. Triangle A1PD1 congruent to B1QC1 ( case HA) so rD-rA=rC-rA

Or rA+rC=rB+rD