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

Labels:
circle,
incircle,
transversal,
triangle

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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)