Geometry Problem

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## Friday, March 4, 2011

### Problem 591: Triangle, Incenter, Incircle, Inradius, Tangency Point, Midpoint, Altitude

Labels:
altitude,
congruence,
incenter,
incircle,
inradius,
midpoint,
tangency point,
triangle

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2b.EM

ReplyDelete= 2b.(AM–AE) = b.(AC–2AE)= b.(b–2cCosA)

= b^2–2bcCosA = b^2–(b^2+c^2–a^2)

= a^2–c^2

2b.DM = b.2DM = b(a–c)from Problem 587

r:EF=DM:EM=b(a – c):(a^2 – c^2)=b:(a + c )

b.EF=r.a + c)

b.BF=b.(BE-EF)=b(cSin A)–r(a+c)

=2Δ–r(a+c)=2rs–r(a+c)=r(2s–a–c)=rb

BF=r follows

Let 2p=a+b+c and h=BE

ReplyDeleteWe have S= Area(ABC)= p.r = ½* b*h

So b=BE= 2.p.r/b = r+r.(a+c)/b (1)

1. Triangle MDO similar to Triangle MEF ( case AA)

So EF=r*(ME/MD) (2)

2. We have a^2-c^2 = CE^2-AE^2=(b/2+ME)^2-(b/2-ME)^2=2.b.ME

So ME=(a^2-c^2)/(2.b)

And MD= (a-c)/2 See Problem 587

Replace value of ME and MD in (2) We have EF=r.(a^2-c^2)/(2.b) *(2/(a-c))= r.(a+c)/b (3)

we have BF=BE-EF= r

Replace value of BE and EF from (1) and (3) to above expression we have BF=r

Peter Tran

(BC+AB).2DM = (BC+AB).(BC-AB)=BC^2–AB^2

ReplyDelete=CE^2–AE^2=(CE+AE).(CE–AE=AC.2EM

EF/r=EF/OD=EM/DM=(BC+AB)/AC=(a+c)/b

So BF=BE–EF=BE–EF

=(2Δ/b)–r(a+c)/b=[2rs-r(a+c)]/b

=r(2s–a–c/b=br/b=r