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