tag:blogger.com,1999:blog-6933544261975483399.post767211962051441436..comments2023-03-25T01:08:45.796-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1024: Contact, Gergonne Triangle, Point on an arc of Incircle, Perpendicular, DistancesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-21233071846343478812017-03-21T19:55:32.210-07:002017-03-21T19:55:32.210-07:00https://goo.gl/photos/8vk1AQ8RWkqjdc3Q9 Let PQ is...https://goo.gl/photos/8vk1AQ8RWkqjdc3Q9<br /><br />Let PQ is the diameter of circle O and R= radius of circle O<br />Define a as ∠ (PTaB)= ∠ (PQTa)<br />. b=∠ (BTcP)= ∠ (PQTc)<br />. c= ∠ (ATbTc)= ∠ (TbQTc)<br />See sketch for details<br />We have<br />Sin(a)=PTa/2.R <br />Sin(b)=PTc/2R<br />Sin(b+c)= PTb/2R <br /> a1=PTa*sin(a)=(PTa)^2/(2R) =&gt; PTa= sqrt(2.R.a1)<br />b1=PTb*sin(b+c)=(PTb)^2/(2R)=&gt; PTb= sqrt(2.r.b1)<br />c1=PTc*sin(b)=(PTc)^2/(2R) =&gt; PTc= sqrt(2.R.c1)<br />Apply Ptolemy’s theorem in quad. PTaTbTc <br />we have TaTc. PTb= PTc.TaTb+ PTa.TbTc<br />Plug these values in above equation we will get the result<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14915264505803280062017-03-16T13:33:17.864-07:002017-03-16T13:33:17.864-07:00Can someone post a solution to this question, I&#3...Can someone post a solution to this question, I&#39;m having a difficult time proving itAnonymoushttps://www.blogger.com/profile/11988214253974032468noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88007811056510984222014-06-21T16:55:48.470-07:002014-06-21T16:55:48.470-07:00Denote inscribed angles of arcs PTa, PTc, and TcTb...Denote inscribed angles of arcs PTa, PTc, and TcTb to be a, b, and c respectively.<br />a1=PTa*sin(a)=(PTa)^2/(2R)<br />b1=PTb*sin(b+c)=(PTb)^2/(2R)<br />c1=PTc*sin(b)=(PTc)^2/(2R)<br />Plugging these into equation to be proved yields Ptolemy&#39;s Theorem in quadrilateral PTaTbTc.<br />Ivan Bazarovnoreply@blogger.com