## Monday, April 13, 2009

### Problem 29. Right Triangle, Altitude, Incircle, Angle bisector, Inradius

Proposed Problem

Level: High School, SAT Prep, College geometry

#### 1 comment:

1. join A and C to O
A + C = 90 => A/2 + C/2 = 45
OBF = 45 - A/2 => OBF = C =>
BOFC cyclic ( OBF = OCF = C/2, angles in the same segm)
=>
OBF = OFB (OFB and OCB = C/2, arc OB )
=> ▲OBF isoceles => OF = r√2 ( OB diagonal of squares )
=> MF = OM = r
in the same way EM = r

=> EF = 2r

=> OEF = OFE = 45° => EOF = 90°
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2) 3) 4) see 1)
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5) EOH = GOH = 90° as vertical angles =>
B + GOH = 180° => BGOH cyclic
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6) BGO + BHO = 180° =>
180 - AGO + BHO = 180 =>
AGO = BHO =>
▲PGO = ▲QOH ( P, Q tg points in AB, BC )
=> GO = OH
=> EO + OH = FO + OG
=> EH = FG
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7) see 6) and 1)
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8)QHO = 45° + C
▲BOH => 45°+ BOH + 45° + C = 180° =>
BOH = A ( C+A=90°)
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9) GOH = 90° & GO = HO => HGO 45° =>
HGO = AFG ( alternate ang ) =>
GH//AC
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10) ▲EGF = ▲EFH
EF common side
EH = FG
OEF = OFE = 45°
=>
EG = FH
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