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Geometry ProblemClick the figure below to see the complete problem 496 about Triangle, Two Squares, 90 Degrees, Concurrency, Angle Bisectors.
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 = 903) see 2)
a correctionsecond row of 2) have to be FPC = FBC
Note that triangles ABG congruence with Tri. DBC ( case SAS) so DC=AGTri. DBC is the image of tri. ABG in the rotational transformation with center at B and rotational angle =90 so DC perpendicular to AGSince angle GPC=90 so quadrilateral FGFC is cyclic with diameter GC and arc GF=arc FC=90Angle 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