Geometry Problem

Click the figure below to see the complete problem 567.

Zoom at: Geometry Problem 567

## Saturday, January 15, 2011

### Problem 567: Right triangle, Incircle, Excircle, Collinear tangency points

Labels:
collinear,
excircle,
incircle,
right triangle,
tangency point

Subscribe to:
Post Comments (Atom)

http://img843.imageshack.us/img843/4514/problem567.png

ReplyDeleteLet 2p=perimeter of tri. ABC ; r= radius of incenter circle

We have CF=p and AG= p-BC

FB=CF-BC=p-BC

So AG=FB

Triangle FBG congruent to tri. AGD ( case SAS)

and angle(FGB)=angle(ADG)=angle(AGH)

So F, G, H are collinear

Peter Tran

EA//GH ( EAH + GHA = 90 + A/2 + 90 - A/2 )

ReplyDeleteEF//AG, EF = AG = R => EFGA paralelogram

=> EA//FG

from G two // to AE => F, G, H collinear

Other,

ReplyDeleteADGH são cocíclicos => DGH=a/2 (i)

M é a interseção de AD com GH, então AMBF tb é cocíclico e BFG = a/2 (ii)

De (i) e (ii) BGF = 90 - a/2 = AGH => F, G, H colineares.

If the circle centered on the point D intersects the BC in K and s is half-perimeter of triangle ABC.(AH=AG=s-a=BF, CH=CK=s-c, BG=BK=s-b).Then CH/HA.AG/GB.BK/KC=1 and from the inverse theorem Menelaus, the points F,Gand H are collinear.

ReplyDeleteAPOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE