Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 768.
First, we have(1) DH = AD = BC = BF(2) DG = CD = AB = BE(3) Angle GDH = 240 - Angle ADC = 240 - Angle ABC = Angle EBFHence, triangle DGH congruent to triangle BEF. Thus, GH = EF. Similarly, we can prove that FG = EH. Hence, EFGH is a parallelogram.
EB=BA=CD=DGAnd BF=BC=AD=DHAnd ∠(EBF)=360-120-∠(ABC)= ∠(GDH)∆ EBF congruence to ∆ GDH ….( Case SAS) => EF=HGSimilarly we have ∆EAH congruence to ∆ GCF => EH=FGSo EFGH is a parallelogram
Let AC ∩ BD = O∆OBF ≡ ∆ODH ∴ ∠BOF = ∠DOH and OF = OHF,O,H are collinear and FH is bisected at O.///ly EG is bisected at O.Hence EFGH is a //ogram
The opposite Tr. s thus formed by the equilateral triangles are congruent SAS since the opposite angles of the original parellelogram are equal. So the opposite sides of these Tr. s are equal. Hence the quadrilateral has opposite sides equal and is thus a parallelogramSumith PeirisMoratuwaSri Lanka