Wednesday, August 8, 2018

Geometry Problem 1375: Isosceles Triangle, Interior Cevian, Exradius, Excircle, Altitude to the Base

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 1375: Isosceles Triangle, Interior Cevian, Exradius, Excircle, Altitude to the Base.

2 comments:

  1. Let the smaller circle be P and the larger Q.
    Let PD meet BE at F.
    Let BD meet circle P at U and circle Q at V. Let AC meet circle P at X and circle Q at Y.
    Let AB=a=BC, AE=EC=d, AX=b and CY=c.

    Now a+b = x+y ............(1)

    Also DV = DY so, y+a-c = d-b-x+d+c
    Therefore c-b = a-d ...(2) where we have used (1) above

    Now EF = dh/(a+d) and so from similar triangles

    b/r1=c/r2=h/(a+d)=(c-b)/(r2-r1)=(a-d)/(r2-r1) from (2) above

    Hence h(r2-r1) = (a+d)(a-d) = a^2-d^2 = h^2

    Therefore r2 - r1 = h

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. https://photos.app.goo.gl/qkDfNLjerBjioR1u9
    Denote AB=BC= a
    EC=EB=b
    CF= x and AL=AH=y
    We have BG=a-x and BH=BM=a+y
    EL=b-y
    So MG=2a-x+y and LF= 2b+x-y
    Equate MG= LF we have b+x=a+y => EF=BH=BM=BN
    We have ∠ (GBC) is an external angle of triangle DBC
    And ∠ (BDC) is and external angle of triangle ABD
    and BO2 and BO1 are angle bisectors of angle GBC and angle HBM
    With angles manipulation we will have ∠ (MBO1) = ∠ (BNO2)
    Triangle BMO1 congruent to BNO2 ( case SAS)
    So NO2=MO1= r1 and r2-r1= NF=BE= h

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