tag:blogger.com,1999:blog-6933544261975483399.post7804594589474594273..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 719: Incenter, Intersecting Circles, Angle, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-72635548398867079592022-12-07T09:41:05.316-08:002022-12-07T09:41:05.316-08:00I'm sure you are long gone, but how do you kno...I'm sure you are long gone, but how do you know that AO' and BO' are equal arcs without assuming that ray CO' bisects <C?JMnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63849201697282148032021-01-16T08:37:03.311-08:002021-01-16T08:37:03.311-08:00AO' = BO', so in cyclic quadrilateral AO&#...AO' = BO', so in cyclic quadrilateral AO'BC, CO' bisects <ACB.<br /><br />Moreover if <ABC = B, then < AO'C = B (in Circle O) and so <ADO' = 90 - B/2<br />Hence < CAD = (90-B/2) - C/2 = A/2<br />Therefore DA bisects <A in Tr. ABC<br /><br />Hence D is the incentre of Tr. ABC<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85514627990234280152012-01-23T18:28:09.708-08:002012-01-23T18:28:09.708-08:00Equal arcs AO’, BO’ of circle (O) subtend equal an...Equal arcs AO’, BO’ of circle (O) subtend equal angles at the circumference.<br />So ∠ACO’ = ∠BCO’ and so<br />CD bisects ∠ACB …… (i)<br />Next ∠DAB =(1/2)∠DO’B referred to circle(O’)<br />=(1/2)∠CO’B =(1/2)∠CAB referred to circle(O) and so<br />AD bisects ∠CAB …… (ii)<br />From (i) and (ii), it follows that <br />D is the incentre of ∆ABC.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com