Geometry Problem

Click the figure below to see the complete problem 496 about Triangle, Two Squares, 90 Degrees, Concurrency, Angle Bisectors.

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Complete Problem 496

Level: High School, SAT Prep, College geometry

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## Monday, August 9, 2010

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Problem 496: Triangle, Two Squares, 90 Degrees, Concurrency

See also:

Complete Problem 496

Level: High School, SAT Prep, College geometry

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Geometry Problem

Click the figure below to see the complete problem 496 about Triangle, Two Squares, 90 Degrees, Concurrency, Angle Bisectors.

See also:

Complete Problem 496

Level: High School, SAT Prep, College geometry

Labels:
90,
angle bisector,
concurrent,
degree,
square,
triangle

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1) a) ▲DBC = ▲ABG ( side, 90+B,side) => DC = AG

ReplyDeleteb) DPA = 45 - CDB, DAP 45 + BAC => DPA = 90°

2) EPA = EDA = 45, (EDPA cyclic)

FPC = PGC = 45, (PGFC cyclic)

APC = 90

3) see 2)

a correction

ReplyDeletesecond row of 2) have to be FPC = FBC

Note that triangles ABG congruence with Tri. DBC ( case SAS) so DC=AG

ReplyDeleteTri. DBC is the image of tri. ABG in the rotational transformation with center at B and rotational angle =90 so DC perpendicular to AG

Since angle GPC=90 so quadrilateral FGFC is cyclic with diameter GC and arc GF=arc FC=90

Angle GPF=angle FPC = 45 ( face 90 degrees arc)

Similarly quadrilateral EDPA is cyclic and angle DPE=angle EPA=45

E, P and F are collinear ( 45+90+45=180) and EF is angle bisector of angle GPC

Peter Tran

Tr. DBC is congruent to Tr. ABG, SAS

ReplyDeleteHence CD = AG

Also because of the congruency,< BDP = < BAP

Hence ADBP is concyclic and so < BPD = < BPG = 45

Also since ADBP is concyclic, < APD = 90 so APDE is concyclic

Hence <APE = < DPE = 45

Similarly <FPG = FPC = 45

It follows that EPC is a straight line and so AG,CD,EF are concurrent

Further EP or EF bisects < GPC

Sumith Peiris

Moratuwa

Sri Lanka