Proposed Problem

See also: Complete Problem 29, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Monday, April 13, 2009

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

Labels:
altitude,
angle bisector,
incircle,
inradius,
right triangle

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join A and C to O

ReplyDeleteA + 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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