tag:blogger.com,1999:blog-6933544261975483399.post8718303206258464577..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1304: Triangle, 120 Degrees, Angle Bisector, Incenter, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-55975053691137239802017-01-08T10:28:35.019-08:002017-01-08T10:28:35.019-08:00Sumith thank you very much.Sumith thank you very much.APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12288742984004773132017-01-07T09:02:30.650-08:002017-01-07T09:02:30.650-08:00Good shortcut ApostolisGood shortcut ApostolisSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82666281807187141542017-01-06T22:52:32.619-08:002017-01-06T22:52:32.619-08:00Problem 1304
Let's say that the EF and AD ...Problem 1304<br /> Let's say that the EF and AD meet at I1 , CF and ED meet at I2.Then the BD is bisector<br />external angle <ABE so the point D is A-excenter of the triangle ABE(<ABE=60=<EBC).Then the ED is bisector of the <BEC.Similar the FE is bisector of the <AEB.So I1 is incenter of the triangle ABE and I2 is incenter of the triangle BEC.Therefore from problem 1303 we have ΑF+2EI+CD=(AF+IE)+(IE+DC)=AE+EC=AC.<br />APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE <br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44098385095433504742017-01-06T22:08:56.603-08:002017-01-06T22:08:56.603-08:00U need not be Einstein to figure out that < AIC...U need not be Einstein to figure out that < AIC = 150.<br /><br />Let X, Y be on AC such that AF = AX & CD=CY.<br /><br />Tr.s AFI & AXI are congruent SAS and so <br />< AIF = < AIX = 30.<br />Similarly < CIY = 30 and so < XIY = 90.<br />Further < FIB = 90-A/2 (considering angles of Tr. AIB) = <CIE, hence < EIY = 60-A/2 and so < EIX = A/2 + 30<br />But < EXI = A/2 + 30 (external angle of Tr. AIX)<br />So E is the Centre of right triangle XIE<br /><br />Therefore AF + CD + 2IE = AX + CY + XY = AC<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88746965418795824702017-01-06T22:01:17.650-08:002017-01-06T22:01:17.650-08:00https://goo.gl/photos/REErnXGV8nZiPM4F7
(new impro...https://goo.gl/photos/REErnXGV8nZiPM4F7<br />(new improved solution)<br />Draw EP//BC <br />Locate points M and N on AC such that AF=AM and CN=CD ( see sketch)<br />We have ∠AIC= 90+1/2.B= 150<br />∠DIC=∠CIN=∠AIM=∠AIF=30 => ∠MIN=90<br />Triangle BEP is equilateral<br />AD bisect angle A => BD/DC=AB/AC=BP/EC=BE/EC …( Thales’s theorem)<br />So ED bisect angle BEC => E is the circumcenter of right triangle MIN<br />AC=AM+MN+NC=AF+2.EI+ CD<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87339322919578871962017-01-06T17:58:08.924-08:002017-01-06T17:58:08.924-08:00https://goo.gl/photos/dB3gitFmAEA66Z6UA
Let EF cu...https://goo.gl/photos/dB3gitFmAEA66Z6UA<br /><br />Let EF cut I at J and EP//AB ( P on BC) See sketch<br />We have ∠ABE=∠EBD=∠BEP=60 => BEP is isosceles=> BE=BP<br />Since CF bisect angle ACB => AF/FB=AC/BC<br />But AC/BC=AE/BP ( Thales’s theorem)= AE/BE<br />So EF bisect angle AEB<br />Per the result of problem 1303 with triangle ABE ( AI and EF bisect angles A and E and angle B= 60)<br />We have AF+IE= AE … (1)<br />Similarly with triangle EBC we have IE+DC=EC.. (2)<br />Add (1) to (2) we have AF+2.EI+DC=AE+EC=AC<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com