Wednesday, June 20, 2012

Problem 768: Parallelogram with Equilateral Triangles on Sides

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 768.

1. First, we have
(1) DH = AD = BC = BF
(2) DG = CD = AB = BE
(3) Angle GDH = 240 - Angle ADC
= 240 - Angle ABC = Angle EBF

Hence, triangle DGH congruent to triangle BEF.
Thus, GH = EF.

Similarly, we can prove that FG = EH.
Hence, EFGH is a parallelogram.

2. EB=BA=CD=DG
And ∠(EBF)=360-120-∠(ABC)= ∠(GDH)
∆ EBF congruence to ∆ GDH ….( Case SAS) => EF=HG
Similarly we have ∆EAH congruence to ∆ GCF => EH=FG
So EFGH is a parallelogram

3. Let AC ∩ BD = O
∆OBF ≡ ∆ODH
∴ ∠BOF = ∠DOH and OF = OH
F,O,H are collinear and FH is bisected at O.
///ly EG is bisected at O.
Hence EFGH is a //ogram

4. 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 parallelogram

Sumith Peiris
Moratuwa
Sri Lanka