tag:blogger.com,1999:blog-6933544261975483399.post7657049003921658381..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 907: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, Tangent, InradiusAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-81698821991529204342013-08-04T16:44:36.914-07:002013-08-04T16:44:36.914-07:00Minor correction due to typo error.
In right tria...Minor correction due to typo error.<br /><br />In right triangle IHA we have IH=x= 3*tan(A/2)<br />in right triangle IGC we have x= CG*tan(C/2)= 1/tan(A/2) ...(A supplement to C)<br />Replace it in 1st expression we have x=3(1/x) => x=SQRT(3)<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35690596064155460872013-08-04T15:28:11.816-07:002013-08-04T15:28:11.816-07:00build CI IG and AI
BAH=2a=>C=180-2a
IAH=a, ICG=...build CI IG and AI<br />BAH=2a=>C=180-2a<br />IAH=a, ICG=90-a<br />CGI=IHA=90=>CIG=a=IAH<br />=>CIG~IAH<br />1/x=x/3=>x=sqrt(3)Anonymoushttps://www.blogger.com/profile/11694885671410014993noreply@blogger.com