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

5 comments:

  1. 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. @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. /_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. 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. 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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