tag:blogger.com,1999:blog-6933544261975483399.post8564964210331938610..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1470: Tangential Quadrilateral, Incircles, Tangent, Parallel, RhombusAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-31908288441896287672020-05-16T22:30:28.584-07:002020-05-16T22:30:28.584-07:00the url you posted doesn't have any imagethe url you posted doesn't have any imageAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23277506629698618492020-04-18T10:17:59.493-07:002020-04-18T10:17:59.493-07:00https://photos.app.goo.gl/jtXfHvuJLwERBHgg7
Define...https://photos.app.goo.gl/jtXfHvuJLwERBHgg7<br />Define point I, F1 to F4 and G1 to G4 as shown on sketch<br />Note that BC, A1C2 and O1O2 concur at I<br />And Oa, Ob, Oc, Od are located at middle of arc T1T4, T1T2, T2T3, T3T4<br />1. We have Angle(BC, ObOc)= ½( Arc OcT2- ArcObT2)= Angle( A2C1, ObOc)... ( 1)<br />And Angle(T1T3,ObOc)= ½(ArcOcT3-Arc(ObT1)… (2)<br />Since Ob and Oc are located at middle of arc T1T2, T2T3 => (1)=(2)<br />And A2C1//T1T3<br />2. Similarly we will have the same result for other 3 sides and E1E2E3E4 is a parallelogram<br />We have AD+BC=AB+CD<br />With manipulation of equal sides we have F2F3+F4F5=G1G2+G3G4<br />With manipulation of equal sides we have E2E3+E1E4=E1E2+E3E4<br />So all 4 sides of E1E2E3E4 are equal and E1E2E3E4 will be a rhombus<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com