tag:blogger.com,1999:blog-6933544261975483399.post6602295169242059443..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1251: Triangle, Circle, Diameter, Perpendicular, 90 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-10869437973067035352016-08-22T22:12:14.772-07:002016-08-22T22:12:14.772-07:00Note that GBDF, AEDF and BEFH are all cyclic quadr...Note that GBDF, AEDF and BEFH are all cyclic quadrilaterals. <br /><br />Hence < GBF = < GDF = < AEF = < BHF<br /><br />So < GBH = < GBF + < HBF = < BHF + <HBF = 90<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1640272239054620202016-08-22T15:53:00.311-07:002016-08-22T15:53:00.311-07:00https://goo.gl/photos/4guoy5kN3aqGP59PA
Draw circ...https://goo.gl/photos/4guoy5kN3aqGP59PA<br /><br />Draw circle O1 diameter GB and circle O2 diameter HB<br />Note that circle O1 pass through points F and D<br />Circle O2 pass through points F and E<br />We have ∠( GBF)= ∠(GDF)= x<br />∠(FBH)= ∠(FEH)=y<br />But ∠(FEC)= ∠(FDC)=y<br />But ∠(GDF)+ ∠(FDC)=90= x+y<br />So ∠ (GBH)= ∠(GBF)+ ∠(FBH)=x+y=90<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com