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

Labels:
altitude,
incenter,
incircle,
right triangle

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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