Geometry Problem

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Level: High School, SAT Prep, College geometry

## Thursday, December 30, 2010

### Problem 561: Triangle, Cevian, Incenter, Perpendicular, 90 Degrees

Labels:
90,
angle,
cevian,
degree,
incenter,
incircle,
perpendicular,
right angle,
triangle

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Call the radii of circles (E) and (F) as r1 and r2

ReplyDeleteDivide the segment EF in the ratio r1:r2 at K.

Through K draw the other (transverse)common tangent LN.

GE, GF bisect the supplementary angles MGN and NGH

Hence angle EGF is a right angle

http://img84.imageshack.us/img84/9576/problem561.png

ReplyDeleteLet I is the midpoint of EF and M is the projection of I over AC

We have angle(EDF)=90 ( see problem 559) and HD=GK (See problem 560)

M is the midpoint of HK . Since HD=GK so M is also the midpoint of DG and triangle DIG is isosceles.

And IE=IF=ID=IG .

Quadrilateral EDGF is cyclic with center of circumcircle at I and angle( EGF)= 90

Peter Tran

To Pravin

ReplyDeleteYour solution assume that L,N and G are collinear.

It is not clear to me .Please explain .

Peter