Sunday, August 5, 2018

Geometry Problem 1374: Isosceles Triangle, Exterior Cevian, Incircle, Excircle, Tangency Points, Parallel Lines

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 1374: Isosceles Triangle, Exterior Cevian, Incircle, Excircle, Tangency Points, Parallel Lines.

Posted by Antonio Gutierrez at 2:16 PM
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2 comments:

  1. Sumith PeirisAugust 6, 2018 at 9:17 AM

    Let the altitude BKL of Tr. ABC meet AC at L and the parallel to AC thro E at K.

    Let BL = h, AB = BC = a, BF = a - b, CH = c, AL = LC = d

    Let < FBE = < FKE = x, BFEK being cyclic
    Let < KHL = y

    As per my solution to Problem 1373, BK = h - r1 = r2 = GH
    So BKHG is a parallelogram and so KH //BG

    Now tan x = r1/(a-b) and tan y = r1/(c + d)

    But since a - b = c + d again as per problem 1373, tan x = tan y
    Hence x = y
    So FKH must be collinear since EK // AH

    So finally FKH // BG

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Sailendra TAugust 14, 2018 at 6:52 AM

    Drop a Perpendicular from B to AC and meet at P
    Drop perpendicular from E to BP and meet at Q
    Join BE and let m(ABE)=m(EBF)=@ and m(BAC)=m(BCA)=$ => m(GBC)=$-@ -------------(1)
    Connect QH and form the right tr. QPH
    In Right tr.QPH, QP=EF (radius of small circle) ------(2)
    PH=BF------------(3)
    Hence QPH and EFB are congruent (SAS)
    =>m(QHP)=@ ---------------(4)

    Since m(EFB)=m(EQB)=90 => BFEQ are concylic
    =>m(FBE)=m(FQE)=@ ---------------(5)

    From (4) and (5), F,Q and H are collinear
    Simple angle chasing in tr.BDA gives m(BDA)=m(FDH)=$-2@
    Hence, m(HFD)=180+@-$------------------(5)
    We know m(GBD)=180-m(GBC)=180-$+@---------------(6)

    From (5) and (6) BG//HF

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