Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
https://goo.gl/photos/dB3gitFmAEA66Z6UALet EF cut I at J and EP//AB ( P on BC) See sketchWe have ∠ABE=∠EBD=∠BEP=60 => BEP is isosceles=> BE=BPSince CF bisect angle ACB => AF/FB=AC/BCBut AC/BC=AE/BP ( Thales’s theorem)= AE/BESo EF bisect angle AEBPer the result of problem 1303 with triangle ABE ( AI and EF bisect angles A and E and angle B= 60)We have AF+IE= AE … (1)Similarly with triangle EBC we have IE+DC=EC.. (2)Add (1) to (2) we have AF+2.EI+DC=AE+EC=AC
https://goo.gl/photos/REErnXGV8nZiPM4F7(new improved solution)Draw EP//BC Locate points M and N on AC such that AF=AM and CN=CD ( see sketch)We have ∠AIC= 90+1/2.B= 150∠DIC=∠CIN=∠AIM=∠AIF=30 => ∠MIN=90Triangle BEP is equilateralAD bisect angle A => BD/DC=AB/AC=BP/EC=BE/EC …( Thales’s theorem)So ED bisect angle BEC => E is the circumcenter of right triangle MINAC=AM+MN+NC=AF+2.EI+ CD
U need not be Einstein to figure out that < AIC = 150.Let X, Y be on AC such that AF = AX & CD=CY.Tr.s AFI & AXI are congruent SAS and so < AIF = < AIX = 30.Similarly < CIY = 30 and so < XIY = 90.Further < FIB = 90-A/2 (considering angles of Tr. AIB) = <CIE, hence < EIY = 60-A/2 and so < EIX = A/2 + 30But < EXI = A/2 + 30 (external angle of Tr. AIX)So E is the Centre of right triangle XIETherefore AF + CD + 2IE = AX + CY + XY = ACSumith PeirisMoratuwaSri Lanka
Problem 1304 Let's say that the EF and AD meet at I1 , CF and ED meet at I2.Then the BD is bisectorexternal 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.APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE
Good shortcut Apostolis
Sumith thank you very much.