See complete Problem 64
Triangle, Incircle, Transversal. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 64
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See complete Problem 64
Triangle, Incircle, Transversal. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
By Menelaus's theorem, AG/GC*CE/EB*BD/DA = -1
ReplyDeleteRemembering that CE=CF=e,DB=BE and FA=AD=d. we've (x+d+e)/x * e/EB *EB/d = 1. Hence,
(x+d+e)/x * e/d = 1 or x = -e^2-ed/(e-d)
= e(d+e)/(d-e)
QED.
Ajit: ajitathle@gmail.com
We see that AD = d and that CE =e.
ReplyDeleteAlso :
∠BDE = ∠BED = ∠GEC= t
and
∠ADG=180-t
∠CGE = u
than we use the sine rule on triangle ECG :
sin(t)/x = sin(u)/e
sin(t)/sin(u) = x/e (1)
and using the sine rule on triangle ADG and the fact that sin(180-t)=sin (t)
sin(180-t)/(x+d+e) = sin(u)/d
sin(t)/sin(u) = (x+d+e)/d (2)
From equation 1 and 2 we get :
x/e = (x+d+e)/d
xd = xe+e(d+e)
x(d-e) = e(d+e)
x = e(d+e)/(d-e)
Draw CI parallel to AB to intersect AG in I (see graph here :
ReplyDeletehttps://drive.google.com/file/d/1O5LQpv5h_kdsRi72etJfKxBlp8v3sJem/view?usp=sharing ).
As O incenter of ΔABC and D, E and F tangency points of the incircle to ΔABC :
- ΔOCF and ΔOCE are similar with common side OC : CE = CF = e
- ΔOAD and ΔOAF are similar with common side OA : AD = AF = d
- ΔOBD and ΔOBE are similar with common side OB : BD = BE and ΔDBE is isocele in B.
Also : e.(d+e+x) = d.x
=> e.(d+e) = x.(d-e) => x = e.(d+e)/(d-e) QED
Missing part after "Also" : <CIG = <ADG by construction of I so <CIE = <BDE = <DEB = <CEI therefore ΔECI isocele in C and CI = CE = e.
ReplyDeleteLastly : In ΔADG: CI//AD therefore CI/AD = CG/AG translates into e/d = x/(d+e+x) => e.(d+e+x) = d.x => e.(d+e) = x.(d-e) => x = e.(d+e)/(d-e) QED
ReplyDeleteHello Antonio, what is the max number of characters allowed. My earlier messages were truncated I don't know why ?
ReplyDeleteHello Greg,
Delete4,096 characters per comment.