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.
Thursday, October 2, 2008
Elearn Geometry Problem 186
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PROOF of PROBLEM 186
ReplyDeleteM – 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
Let m(A)=A => m(C)=90-A
ReplyDeleteExtend 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
Let X and Y be the tangency points of the 2 circles on AC of radius r1 (Tr.ABD) and r2 (Tr.CBD). Let BD touch circle r1 at V and circle r2 at U.Let BC touch circle r2 at W
ReplyDeleteEVDX and FUDY are squares and so < EDF = 90
According to my solution for Problem 23, r1 = rc/b and r2 = ra/b (r=inradius of ABC)
So in right triangle EDF, DF/DE = r2/r1 = a/c.
Hence Tr.s ABC and EDF are similar and < DEG = A
Angle chasing around E gives < BEG = 90-A/2 and since < EBG = 45+A/2, < BGE = 45 = < CFD
So CDFG is concyclic and < CDG = CFG = 45 - (45-A/2) = A/2
Hence < BDG = 90- A/2 = < DGB since in Tr. BDG, < DBG = A
Therefore BD = BG
Sumith Peiris
Moratuwa
Sri Lanka
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