Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 96Similar Triangles, Incenters, Parallelogram. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
join B to O1 & O extend to P ( P on AC )(same bisector)extend GO3 to Q ( Q on AC )ang APB = 1/2 B + Cang MQG = 1/2 G + C ( AB//MF => ang B = ang G )=>ang APB = ang MQG=> PB // QO2ang DEA = ang C ( DE // AC )=>1/2 DEA = 1/2 C=>O2O1 // CO
To Antonio:The written enunciate of problem 96 is a repetition of problem 94. The figure is all right.
Thanks Nilton. The written enunciate of problem 96 has been updated.
Alternative solution to problem 96.Points O, O1 and B are collinear, on the bisector of ang(ABC). Points O2, G and O3 are collinear, on the bisector of ang(MGC). As triangles ABC, DBE, FGE and MGC are similar,then OB/BC = O1B/BE = O2G/EG = O3G/GC = (O1B+O2G+O3G)/(BE+EG+GC).But BE + EG + GC = BC, so OB = O1B + O2G + O3G, which is the same as OB – O1B = O2G + O3G. Besides OO1 = OB – O1B = O2G + O3G = O2O3.Being ABC and BGC similar, thenang(ABO) = ang(MGO3), and, as AB // MR, thenOO1 // O2O3.We can see now that OO1O2O3 is a parallelogram because it has two opposite sides equal and parallel.