Thursday, October 2, 2008

Elearn Geometry Problem 186

Right Triangle, Altitude, Incenter

See complete Problem 186 at:
www.gogeometry.com/problem/p186_right_triangle_circle.htm

Right Triangle, Altitude, Incenters, Circles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. PROOF of PROBLEM 186
    M – tangency point on BC, K - tangency point on BD
    angleDBC=angleBAD=α→
    angleDBF=angleEAD=∝/2→
    angleADE=angleBDF=45°→
    ∆AED≈∆BFD→ ED/DF=AD/BD,angleEDF=angleADB=90°→∆ABD≈∆EFD→angleEFD=β
    angleBFE=180°-β-∝/2-45°= ∝/2+angleBGF→
    angleBGF=45°,
    angleFKB=angleFMG=90°→MG=KD=R→
    BM=BK,BG=BM+R=BK+R=BD

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  2. Let m(A)=A => m(C)=90-A
    Extend GFE to meet AB at H
    Join BE,ED and form the triangle BED (with Angles 45-A/2,90+A/2,45)
    Join BF,FD and form the triangle BFD (with Angles A/2,135-A/2,45)
    Join AE and form the triangle AED (A/2,135-A/2,45)
    Join FC and form the triangle DFC (45,90+A/2,45-A/2)
    A bit of angle chasing, we can observe that the triangles ABE and BCF are similar (AAA)
    => AB/BC=AE/BF ----------(1)
    Similarly the triangles AED and BFD are similar
    => AE/BF=ED/FD ----------(2)
    From(1) and (2), the triangles ABC and EDF are similar => m(DEF)=A, hence m(BEG)=90-A/2

    Consider the triangle BEG, we can derive that m(BGE)=45 (since m(EBG)=45+A/2 and m(BEG)=90-A/2)
    Similarly, in the triangle BFH , m(BHF)=45
    => Triangle BGH is an isosceles right angle triangle -------------(3)

    Since m(AHE)=135 and m(ADE)=45 => AHED is concyclic, similarly CGFD is concyclic
    => m(HDE)=m(HAE)=A/2 and m(FDG)=m(FCG)=45-A/2
    => m(HDG)=m(HDE)+m(EDB)+m(BDF)+m(FDG)= 135 = 1/2(360-90)=1/2(360-m(HBG)) -----------(4)
    => From (3) and (4), B is the Circumcenter of the triangle HDG and hence BG=BH=BD

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