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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Online Geometry theorems, problems, solutions, and related topics.

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