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.

## Thursday, April 16, 2020

### Geometry Problem 1470: Tangential Quadrilateral, Incircles, Tangent, Parallel, Rhombus

Labels:
circumscribed,
common tangent,
Geometry,
incircle,
ipad,
ipadpro,
parallel,
poster,
rhombus,
tangential quadrilateral,
typography

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https://photos.app.goo.gl/jtXfHvuJLwERBHgg7

ReplyDeleteDefine point I, F1 to F4 and G1 to G4 as shown on sketch

Note that BC, A1C2 and O1O2 concur at I

And Oa, Ob, Oc, Od are located at middle of arc T1T4, T1T2, T2T3, T3T4

1. We have Angle(BC, ObOc)= ½( Arc OcT2- ArcObT2)= Angle( A2C1, ObOc)... ( 1)

And Angle(T1T3,ObOc)= ½(ArcOcT3-Arc(ObT1)… (2)

Since Ob and Oc are located at middle of arc T1T2, T2T3 => (1)=(2)

And A2C1//T1T3

2. Similarly we will have the same result for other 3 sides and E1E2E3E4 is a parallelogram

We have AD+BC=AB+CD

With manipulation of equal sides we have F2F3+F4F5=G1G2+G3G4

With manipulation of equal sides we have E2E3+E1E4=E1E2+E3E4

So all 4 sides of E1E2E3E4 are equal and E1E2E3E4 will be a rhombus

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