## Monday, August 9, 2010

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

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

Complete Problem 496

Level: High School, SAT Prep, College geometry

1. 1) a) ▲DBC = ▲ABG ( side, 90+B,side) => DC = AG
b) DPA = 45 - CDB, DAP 45 + BAC => DPA = 90°
2) EPA = EDA = 45, (EDPA cyclic)
FPC = PGC = 45, (PGFC cyclic)
APC = 90
3) see 2)

2. a correction
second row of 2) have to be FPC = FBC

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

4. Tr. DBC is congruent to Tr. ABG, SAS
Hence 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