Tuesday, August 5, 2008

Elearn Geometry Problem 156



See complete Problem 156
Triangle, Circumradius, Exradius, Chord, Secant line. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 10:53 AM
Labels: , , , , , , ,

6 comments:

  1. c .t . e. oFebruary 12, 2010 at 12:59 PM

    I have first half of solution, but on the second part
    using similarity of tr AOE, (OE perpendicular to AD)
    and tr PEG, ( P AD meet BC, G tg point of E to BC )
    I get 2R∙r = AD∙PE, ( ang EAO = DEG = 90 -x - C )
    Please verify

    ReplyDelete
  2. Antonio GutierrezFebruary 12, 2010 at 2:29 PM

    To c.t.e.o:
    Your conclusion 2R.r1 = AD.PE is OK.
    Other conclusions:
    AD.PE = AE.DE = 2R.r1

    ReplyDelete
  3. c .t . e. oFebruary 12, 2010 at 3:52 PM

    x = A/2, T tg point of E to AC

    DEG = 90 - x - C ( first comment )
    DEC = DEG + GEC
    DEC = 90 - x - C + C/2 ( CE bisector of GCT )

    DEC = 90 - x - C/2 (1)

    ECT = ECG = 90 - C/2
    DCE = GCE - GCD

    DCE = 90 - C/2 - x (2)
    ( GCD = x ,have same arc DC as A/2)

    from (1) & (2)
    DEC = DCE
    =>

    DE = DC
    ------------------------------------------
    ▲ADC ~ ▲PCD ( A/2 = PCD = x , see above)

    CD/AD = PD/CD
    DE/AD = PD/DE ( CD=DE)
    (AE - AD)/AD = (PE - DE)/DE
    AE/AD - 1 = PE/DE - 1
    AE/AD = PE/DE

    AE∙DE = AD∙PE = 2R∙r (see first comment)
    -----------------------------------------

    ReplyDelete
  4. AnonymousJuly 7, 2010 at 4:43 AM

    http://ahmetelmas.wordpress.com/2010/05/15/geo-geo/

    ReplyDelete
  5. Nilton LapaJuly 5, 2012 at 4:42 PM

    Solution to problem 156.
    1) The bisector of ang(ABC) meets AE at point I, and the circumcircle at M. Arcs AM and MC are equal. As AE is bisector of ang(BAC) then arcs BD and CD are equal. BE is external bisector, so ang(EBM) = 90º.
    We have ang(DBE) = 90º - ang(MBD) = 90º - arc(MD)/2.
    Besides ang(DEB) = 90º - ang(BID) = 90º - (arc(AM) + arc(BD))/2 = 90º - (arc(MC) + arc(CD))/2 = 90º - arc(MD)/2 = = ang(DBE). Hence BDE é isosceles, with BD = DE.

    2) The extension of BO meets the circumcircle at F, and ang(BDF) = 90º. Let T be the tangent point of the excircle with AC. We have ang(BFD) = ang(BAD) = ang(EAT). So triangles BDF and ETA are similar, with AE/FB = ET/BD. Therefore AE/(2R) = r1/DE and AE.DE = 2Rr1.

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  6. AnonymousSeptember 14, 2024 at 4:38 AM

    Let BO extended meet the Circle O at F
    Let AB extended be tangent at U to the Excircle

    (1) Since BE bisects < DBU,
    < UBE = 90 - B/2 = A/2 + C/2

    Hence < AEB = < UBE - < BAE = C/2 since AE bisects < BAC
    But < BDA = C, so < DBE = C - C/2 = C/2
    Therefore BD = DE

    (2) Now Triangles BDF & AUE are similar (each Triangle having 90 & A/2)
    So BD / BF = UE / AE
    So DE / 2R = r1 / AE
    and AE.DE = 2Rr1

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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