Friday, January 6, 2017

Geometry Problem 1304: Triangle, 120 Degrees, Angle Bisector, Incenter, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1304: Triangle, 120 Degrees, Angle Bisector, Incenter, Measurement.

Posted by Antonio Gutierrez at 3:52 PM
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6 comments:

  1. Peter TranJanuary 6, 2017 at 5:58 PM

    https://goo.gl/photos/dB3gitFmAEA66Z6UA

    Let EF cut I at J and EP//AB ( P on BC) See sketch
    We have ∠ABE=∠EBD=∠BEP=60 => BEP is isosceles=> BE=BP
    Since CF bisect angle ACB => AF/FB=AC/BC
    But AC/BC=AE/BP ( Thales’s theorem)= AE/BE
    So EF bisect angle AEB
    Per 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

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  2. Peter TranJanuary 6, 2017 at 10:01 PM

    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=90
    Triangle BEP is equilateral
    AD 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 MIN
    AC=AM+MN+NC=AF+2.EI+ CD

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  3. Sumith PeirisJanuary 6, 2017 at 10:08 PM

    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 + 30
    But < EXI = A/2 + 30 (external angle of Tr. AIX)
    So E is the Centre of right triangle XIE

    Therefore AF + CD + 2IE = AX + CY + XY = AC

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. APOSTOLIS MANOLOUDISJanuary 6, 2017 at 10:52 PM

    Problem 1304
    Let's say that the EF and AD meet at I1 , CF and ED meet at I2.Then the BD is bisector
    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.
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

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