tag:blogger.com,1999:blog-6933544261975483399.post7963000223223834016..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 291: Triangle, Circle, Radius, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-50920924578321440382021-03-06T07:22:05.210-08:002021-03-06T07:22:05.210-08:00BEDF are concyclic and <EDF = <ABC therefore...BEDF are concyclic and <EDF = <ABC therefore e, the chord of <EDF in circle BEDF, is to b, the chord of <ABC in circle ABC, what the radius of BEDF is to R i.e. e/b = (d/2)/R QED Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63628000314870707122019-01-04T13:45:19.750-08:002019-01-04T13:45:19.750-08:00Sumith Peiris' answer:
Drop perpendiculars FP...Sumith Peiris' answer:<br /><br />Drop perpendiculars FP (=p) to BC and OQ to AC. Let FB = q<br />From similar Triangles, <br />b/2R = p/q. and e/d = p/q and the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74642290522762292452018-08-22T03:00:51.956-07:002018-08-22T03:00:51.956-07:00BEDF is cyclical and BD is the diameter of the cir...BEDF is cyclical and BD is the diameter of the circle.<br />Let M be the midpoint of BD then M is the center of the circle BEDF.<br />Let N be the center of the circumcircle of ABC.<br /><br />Draw the circle of M.<br />Join FM and extend FM to meet the circle M in the point P.<br />Join P with E.<br /><br />Join AN and extend AN to meet the circle N in the point Q.<br />Join Q with C<br /><br />We can see that ∠ABC=∠FDE and ∠ABC =∠AQC.<br />Also ∠FDE =∠FPE therefore ∠ABC=∠FPE<br /><br />Because FPE and AQC are right angled triangles they are similar and from similarity we get:<br /><br />AC/AQ=FE/FP<br />b/2R=e/d<br />bd=2Re<br />Arifhttps://www.blogger.com/profile/16774426807135540272noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67006604487737697052014-01-26T22:12:03.744-08:002014-01-26T22:12:03.744-08:00Apply sine law in triangle ABC=> sin(B)= b/2R
A...Apply sine law in triangle ABC=> sin(B)= b/2R<br />Apply sine law in triangle EDF => sin(EDF)= e/d<br />Since EDFB is cyclic quadrilateral so angle (B)= angle (EDF)<br />So we have b/2R= e/d => 2R.e=bd<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31438025835932141432013-02-25T03:19:42.132-08:002013-02-25T03:19:42.132-08:00Correction: Let O be the circum-centre of Tr. ABC....Correction: Let O be the circum-centre of Tr. ABC.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72564654988342273982013-02-24T19:00:01.304-08:002013-02-24T19:00:01.304-08:00Let M be the mid-point of BD & O the circumcir...Let M be the mid-point of BD & O the circumcircle of Tr. ABC. Then clearly, FM=DM=d/2 and /_FMB = 2*/_FDE=2*/_B since F,D,E & B are concyclic. Further, Tr. FME is isosceles. Likewise, Tr. OAC is isosceles with an apex /_ of 2*B. Hence, Tr. FME /// Tr. OAC which gives us, FM/FE = OA/AC or d/(2e) = R/b or R = bd/(2e) Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78536752321422302582009-05-24T20:28:03.708-07:002009-05-24T20:28:03.708-07:00In quad. DFBE we've DF & DE perpendicular ...In quad. DFBE we've DF & DE perpendicular BF & BE respectively. Hence DFBE is concyclic. Therefore, angle ABC = angle EDF = let's say, B. Now let angle EDB = B1 & angle BDF = B2. Hence, sin(B) = sin(B1+B2) = sinB1*cosB2 + oosB1*sinB2 = (BE/d)(DF/d) + (DE/d)(BF/d) = (BE*DF+DE*BF)/d^2<br />However, by Ptolemy's Theorem, (BE*DF+DE*BF =d*e<br />Thus, sin(B)= d*e/d^2 = e/d ---------(1)<br />Now area of Tr. ABC = acsin(B)/2. Hence, it's circumradius R = abc/4(acsin(B)/2) = b/2sin(B) or R = b/2(e/d) using equation (1). This gives:<br />2R*e = b*d<br />Vihaan Uplenchwar, Dubai: vihaanup@gmail.comVihaanhttps://www.blogger.com/profile/14122321143712979648noreply@blogger.com