Geometry Problem

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

Click the figure below to see the complete problem 657.

## Sunday, August 14, 2011

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

Labels:
incircle,
parallel,
tangency point,
triangle

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

ReplyDeleteAjit

http://img143.imageshack.us/img143/1590/problem657.png

ReplyDeleteConnect 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

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

ReplyDeleteChinese version

Let JK meet AE at L

ReplyDeleteSince 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