Geometry Problem

Click the figure below to see the complete problem 609.

## Sunday, May 29, 2011

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Sunday, May 29, 2011

###
Problem 609: Triangle, Perpendiculars, 90 Degrees, Collinearity

Online Geometry theorems, problems, solutions, and related topics.

Labels:
collinear,
perpendicular,
triangle

Subscribe to:
Post Comments (Atom)

Let(XYZ) denote the area of triagle XYZ.

ReplyDeleteAG/GC

=(ADG)/(GDC)=(AD.GD.sin ADG)/(GD.CD.sin GDC)

=(AD/CD)(sin ADG /sin GDC)

CF/FB

=(CDF)/(FDB)=(CD.FD.sin CDF)/(FD.BD.sin FDB)

=(CD/BD)(sin CDF / sin FDB)

BE/EA

=(BDE)/(EDA)=(BD.ED.sin BDE)/(ED.AD.sin EDA)

=(BD/AD)(sin BDE / sin EDA)

From perpendicularity conditions,

sin ADG = sin FDB, sin GDC = sin BDE, and

sin CDF = sin EDA

Follows

(AG/GC)(CF/FB)(BE/EA)= 1 numercally

Applying Converse of Menelau's Theorem,

E,F,G are collinear