Geometry Problem

Click the figure below to see the complete problem 537 about Rectangle, Midpoints, Diagonal, Angle Bisector.

Go to Complete Problem 537

## Friday, November 5, 2010

### Problem 537: Rectangle, Midpoints, Diagonal, Angle Bisector

Labels:
angle bisector,
diagonal,
midpoint,
rectangle

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Let a=BE=AF, b=AB=EF and c=AG

ReplyDeleteLet K,M, Q are the projection of H over EF, FD and BA

See link below for details

http://img444.imageshack.us/img444/1374/problem537.png

1 Triangle HNF similar to Triangle HBG ( case AA)

We have HK/HQ=b/(2.(b+c))

HQ/2(b+c)=HK/b =a/(2c+b)

So HK= a.b/(2c+b)

2 Triangle FHM similar to triangle FAG ( case AA)

We have HM/c=FM/a so HM=c/a*FM=b.c/(2c+b)

EK=b-HM=b.(b+c)/(2c+b)

3 Note that ratio HK/GP=EK/EP=b/(2c+b)

Triangle EPG similar to triangle EKH ( case SAS)

Corresponding angles GEF will congruence with angle KEH and EF is the angle bisector of angle GEH

Peter Tran

Let be N intersection of AB and HE (extended)

ReplyDeleteFH/GF = OF/BG, EH/NE = OE/BN => BG = BN

=> EN = GE

EH/EN = FH/GF => EH/EG = FH/GF

EH/FH = EG/GF

Let J intersection of CD and EH extended.

ReplyDeleteLet I intersection of CD extended and GH.

Tr. EFH similar to Tr. JIH.

BH is median of EFH and JIH -> AG = ID = DJ.

Then EJ = EG and Tr. EGJ is isosceles and EF is altitude -> EF is bisector of GEH.