tag:blogger.com,1999:blog-6933544261975483399.post383679946536918209..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 906: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-21165207248238883782013-08-03T17:58:00.532-07:002013-08-03T17:58:00.532-07:00Let EG and FH meet at K.
Then
∠FKG
= 1/2 ( ∠FIG ...Let EG and FH meet at K. <br /><br />Then<br />∠FKG<br />= 1/2 ( ∠FIG + ∠EIH )<br />= 1/2 ((180 - ∠FCG ) + (180 - ∠EAH))<br />= 90 °Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11373443434877427842013-08-03T14:55:53.545-07:002013-08-03T14:55:53.545-07:00A=a
C=180-a
FHA=BFH=b [=FGH]
AEG=EGD=c [=EFG]
FH m...A=a<br />C=180-a<br />FHA=BFH=b [=FGH]<br />AEG=EGD=c [=EFG]<br />FH meet EG at K<br />EKH=360-a-b-c=FKG<br />CGK=180-EGD=180-c<br />CFK=180-b<br />360=180-a+180-b+180-c+360-a-b-c<br />a+b+c=270<br />FKE=90<br />Anonymoushttps://www.blogger.com/profile/11694885671410014993noreply@blogger.com