tag:blogger.com,1999:blog-6933544261975483399.post6074851650557723902..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 637: Semicircle, Diameter, Perpendicular, Inscribed Circle, Chord, Tangent, ArbelosAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-21832853815896115602015-11-06T23:21:18.957-08:002015-11-06T23:21:18.957-08:00Using my proof and the result of Problem 636 AG is...Using my proof and the result of Problem 636 AG is a tangent to circle CGB, hence AG^2 = AC.AB<br /><br />But AE^2 also = AC.AB since AE is a tangent to circle BCE<br /><br />So AE = AG<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84387858520953725042011-07-19T21:14:09.879-07:002011-07-19T21:14:09.879-07:00Observe that AEB is a right triangle ( AB is a di...Observe that AEB is a right triangle ( AB is a diameter)<br />so AE^2= AC.AB (1) ( Relation in a right triangle)<br />In circle center D we also have<br />AG^2= AC.AB (2) ( power of point A to circle D)<br />Compare (1) and (2) we get AE=AG<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40711112540342606972011-07-19T20:30:49.992-07:002011-07-19T20:30:49.992-07:00AG^2=AD^2 - DG^2
=(2R-r)^2 - r^2
=4R^2 - 4Rr
=4R(R...AG^2=AD^2 - DG^2<br />=(2R-r)^2 - r^2<br />=4R^2 - 4Rr<br />=4R(R-r)<br />AE^2=AC.AB (tr AEC and tr ABE are similar)<br />=(2R-2r).(2R)<br />=4R(R-r)<br />AE^2=AG^2,<br />AE=AGPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com