## Sunday, August 14, 2011

### Problem 657: Triangle, Incircle, Tangency Points, Parallel Lines

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 657.

1. See problem # 328. J & K are midpoints of AB & AC resply. Hence JK = BC/2 =(BE+CE)/2=(BD+CF)/2 = 4.5
Ajit

2. http://img143.imageshack.us/img143/1590/problem657.png
Connect DF and AE
1. We have (DEF)=(ADF) both angles look the same arc DF
but (AHF) supplement to (DEF)
So (AHF) supplement to ( ADF) >> A,D,F, H is cyclic
Similarly we also have A,G,D,F cyclic
5 points A,G,D,F,H are cyclic ( see picture)
2. We have (EDF)= (FEC) both angles look the same arc EF
And ( EDF)=(GHF) both angles look the same arc GF
So ( FEC)=(GHF) and GH //BC

3. Parallelogram AGEH have diagonals bisect each other so AJ/AB= ½ and KJ= 4.5
Peter Tran

3. http://imgsrc.baidu.com/forum/pic/item/9c16fdfaaf51f3debc44053b94eef01f3b2979fe.jpg

Chinese version

4. Let JK meet AE at L

Since AGEH is a parellelogram AL = LE. But LK//EC so from mid point theorem LK = EC/2 = 2

Similarly JL = BE /2 = 2.5

So JK = 4.5

Sumith Peiris
Moratuwa
Sri Lanka