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

skip to main |
skip to sidebar
## Monday, August 9, 2010

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

See also:

Complete Problem 496

Level: High School, SAT Prep, College geometry

## Link List

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

Subscribe to:
Post Comments (Atom)

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