Geometry Problem

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

Click the figure below to see the complete problem 768.

## Wednesday, June 20, 2012

### Problem 768: Parallelogram with Equilateral Triangles on Sides

Labels:
equilateral,
parallelogram,
side,
triangle

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First, we have

ReplyDelete(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.

EB=BA=CD=DG

ReplyDeleteAnd BF=BC=AD=DH

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

Let AC ∩ BD = O

ReplyDelete∆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

The opposite Tr. s thus formed by the equilateral triangles are congruent SAS since the opposite angles of the original parellelogram are equal.

ReplyDeleteSo 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