Geometry Problem
Click the figure below to see the complete problem 577.
Zoom at: Geometry Problem 577
Wednesday, February 2, 2011
Problem 577: Isosceles triangle, Midpoint, Perpendicular, Concyclic points, Circle
Subscribe to:
Post Comments (Atom)
Join B to D. Since Tr. ABC is isosceles, BD is perpendicular to AC. In Tr. BDC, DE=BD*DC/AC since DE is perpendicular to BC. Now let BD = y and DC = x & let D be the origin. Thus,DE = xy/√x^2+y^2)& E is [xy^2/(x^2+y^2),yx^2/(x^2+y^2)]. Therefore, F is [xy^2/2(x^2+y^2),yx^2/2(x^2+y^2)].while A is (-x,0). Slope BF=[yx^2/2(x^2+y^2) - y]= -(x^2+2y^2)/xy and slope AF= [yx^2/2(x^2+y^2)]/[xy^2/2(x^2+y^2)+x] =xy/(x^2+2y^2) upon simplification. Hence, slope BF*slope AF = -1 or /_AGB is=90 deg. Since ADB and AGB are both rt. angles, we have the necessary result.
ReplyDeleteAjit
http://img7.imageshack.us/img7/2271/problem577.png
ReplyDelete1. Let I is the midpoint of AB.
We have BD perpen. To AC and ID//BC ( Thales theorem)
DE perpen. To BC and ID so DE tangent to circle I, diameter AB
2. Let BF cut circle I at G’ . We will prove that G’ coincide to G
We have BG’ perpen. to AG’ and FE^2=FD^2= FB.FG’ ( power from point F)
So FE/FB=FG’/FE and triangle FEB similar to trị FG’E ( case SAS)
So angle(FG’E)=angle(FEB)= 90 and points A, G’, E are collinear
both G’ and G are the point of intersection of AE and BF so G coincide with G’
Points A,D,G and B are concyclic
Peter Tran