tag:blogger.com,1999:blog-6933544261975483399.post9025738584839048293..comments2023-09-28T06:48:50.134-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1055: Triangle, Altitude, Perpendicular, Area, Circumradius, CircleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-45924162484484843952021-04-19T12:52:57.698-07:002021-04-19T12:52:57.698-07:00Considering usual triangle notations, it can be ob...Considering usual triangle notations, it can be observed that BEFD is cyclic<br />=&gt; m(BED)=m(BFD)=C<br />Hence Tr. BEF similar to BCA<br />=&gt; BE=ad/b<br />Mark O as circumcenter and G as mid-point of BC =&gt; BG=a/2 <br />Observe that Tr.BOG is similar to BFE<br />=&gt;BF=BO.BE/BG<br />=&gt;BF=R.(ad/b)/(a/2)<br />=&gt;BF.b/2=R.d <br />Q.E.DSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68457982107797059982014-12-27T13:05:45.581-08:002014-12-27T13:05:45.581-08:00R=BC/2*BA/BF d=AC*BD/BA Rd=BC*AC*BD/(2BF) Due to s...R=BC/2*BA/BF<br />d=AC*BD/BA<br />Rd=BC*AC*BD/(2BF)<br />Due to similar triangles BC*BD=BF^2<br />Therefore, Rd=AC*BF/2=[ABC]Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44155999597792958052014-11-05T00:06:02.342-08:002014-11-05T00:06:02.342-08:00D,F,E,B are concyclic and they lie on the circle o...D,F,E,B are concyclic and they lie on the circle on BF as diameter.<br />In this circle d = chord EF = (diameter BF) x sin B <br />(i.e.) d = h Sin B where h denotes height BF.<br />In ΔABC we have <br />2Δ= h.b = h. 2R sin B = 2R .h Sin B = 2R.d<br />Hence Δ = R.dPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30050203126623158172014-11-04T21:40:45.297-08:002014-11-04T21:40:45.297-08:00Let S be the area. Let BC=a, AC=b, AB=c. Since Δ...Let S be the area. Let BC=a, AC=b, AB=c. <br /><br />Since ΔABC~ΔDBE, thus DE/AC=BD/AB. <br />d / b = (c sinA sinC) / c<br />d = b sinA sinC = b (a/2R) (c/2R)<br /><br />Hence, S = abc/4R = Rd. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com