Wednesday, September 30, 2009

Problem 361: Right triangle, Incircle, Incenter, Tangency points, Angle

Proposed Problem
Click the figure below to see the complete problem 361 about Right triangle, Incircle, Incenter, Tangency points, Angle.

Problem 361. Right triangle, Incircle, Incenter, Tangency points, Angle.
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Complete Problem 361
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:58 AM
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8 comments:

  1. VihaanOctober 13, 2009 at 9:35 PM

    We've angle AED = angle AED = (180-A)/2 = 90 -A/2
    Futher from Tr. FEC, α = angle AED - angle ACF =90-A/2-C/2 = angle B/2 since angs.(A+B+C)=180 deg. which, in turn, means that α = 90/2 = 45 deg.
    Vihaan: vihaanup@gmail.com

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  2. AnonymousMarch 17, 2010 at 11:53 PM

    @vihaan
    Pls explain this step

    We've angle AED = angle AED = (180-A)/2 = 90 -A/2

    How do we get this?????

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  3. VihaanJuly 10, 2010 at 5:26 AM

    /_AED+/_ADE+/_A=180 but /_AED=/_ADE (equal tangents). So 2/_ADE+/_A=180 or /_ADE=/_AED=90- A/2
    Vihaan

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  4. Sumith PeirisJuly 7, 2015 at 3:34 AM

    Since Tr. ADE is isoceles and OE is perp. to AC < DEO = A/2. But < FCE = 45-A/2 and hence considering the angles of Tr. FEC which add upto 180 we see that alpha= 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. APOSTOLIS MANOLOUDISAugust 12, 2016 at 2:34 PM

    Problem 361
    The ADOE is cyclic (<ADO=90=<AEO) so <EDO=<A/2.Now DO//BC then <FOD=<FCB=<C/2.
    But x=<EFO=<FDO+<DOF=<A/2+<C/2=45.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  6. Benjamin LeisMay 22, 2017 at 12:00 PM

    Draw in the tangents lines and then just angle chase to find <OFE.

    First let G be the intersection of the tangent on BC and x = <OCB

    1. Tr. CGO = Tr.GEO by SSS. So <ECO also equals x.
    2. <COE is 90 -x and <EOF is 90 + x.
    3. DO is parallel to BC so <DOF is also x.
    4. That means <DOE is 90 + 2x and since tr. EDO is isosceles (2 radii) <EDO - <DE0 = 45 - x.
    5. Finally <EFO = 180 - (45 - x + 90 + x) = 45.

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  7. AnonymousMay 24, 2017 at 7:01 AM

    Considering usual triangle notations,Extend CO to meet AB at P
    Since ADE is isosceles, m(ADE) = m(ADF) = 90-A/2
    Also m(OCB) = 45-A/2 => m(CPB) = 45+A/2
    Therefore m(PFD) = m(EFO) = 180-(90-A/2)-(45+A/2) = 45

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  8. MarcoSeptember 1, 2023 at 9:55 PM

    [For easy typing, I use a instead of alpha]
    Let <DAE=x
    AD=AE (tangent property)
    <ADE=90-x/2 (base <s, isos triangle)
    <EDB=180-<ADE (adj, <s on st. line)
    =90+x/2
    <DFC=180-a (adj <s on st. line)
    Consider quadrilateral BCFD
    <FCB=360-<DFC-<FDC-<DBC
    =360-90-(90+x/2)-(180-a)
    =a-x/2--------------(1)

    Consider triangle ABC
    <ACB=90-x (< sum of triangle)
    <FCB=<ACB/2
    =45-x/2-------------(2)

    By equating (1) & (2)
    a=45

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