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Question Number 15784 by mrW1 last updated on 13/Jun/17

Commented by mrW1 last updated on 14/Jun/17

ABCD is a convex quadrilateral. On each side of it a square is constructed. E,F,G,H are center points of these squares. Prove that EG=FH and EG⊥FH.

ABCDisaconvexquadrilateral.Oneachsideofitasquareisconstructed.E,F,G,Harecenterpointsofthesesquares.ProvethatEG=FHandEG⊥FH.

Answered by ajfour last updated on 14/Jun/17

Commented by ajfour last updated on 14/Jun/17

z_1 +z_2 =z_3 +z_4 or z_1 −z_4 =z_3 −z_2 z_H −z_F =((iz_1 )/2)+(z_1 /2)+z_2 −(z_4 /2)+((iz_4 )/2) =((i(z_1 +z_4 ))/2)+(((z_1 −z_4 ))/2)+z_2 =((i(z_1 +z_4 ))/2)+(((z_3 −z_2 ))/2)+z_2 z_H −z_F =((i(z_1 +z_4 ))/2)+(((z_2 +z_3 ))/2) ....(i) z_G −z_E =−((iz_3 )/2)+(z_3 /2)+z_4 −(z_2 /2)−((iz_2 )/2) =−((i(z_2 +z_3 ))/2)+(((z_3 −z_2 ))/2)+z_4 =−((i(z_2 +z_3 ))/2)+(((z_1 −z_4 ))/2)+z_4 =−((i(z_2 +z_3 ))/2)+(((z_1 +z_4 ))/2) . Now (z_G −z_E )×i =(((z_2 +z_3 ))/2)+((i(z_1 +z_4 ))/2) = z_H −z_F ⇒ ∣z_G −z_E ∣ = ∣z_H −z_F ∣ or length EG = length FH and z_H −z_F =e^(iπ/2) (z_G −z_E ) which means line FH is at 90° counterclockwise to EG .

z1+z2=z3+z4orz1−z4=z3−z2zH−zF=iz12+z12+z2−z42+iz42=i(z1+z4)2+(z1−z4)2+z2=i(z1+z4)2+(z3−z2)2+z2zH−zF=i(z1+z4)2+(z2+z3)2....(i)zG−zE=−iz32+z32+z4−z22−iz22=−i(z2+z3)2+(z3−z2)2+z4=−i(z2+z3)2+(z1−z4)2+z4=−i(z2+z3)2+(z1+z4)2.Now(zG−zE)×i=(z2+z3)2+i(z1+z4)2=zH−zF⇒∣zG−zE∣=∣zH−zF∣orlengthEG=lengthFHandzH−zF=eiπ/2(zG−zE)whichmeanslineFHisat90°counterclockwisetoEG.

Commented by mrW1 last updated on 14/Jun/17

very smart! thanks!

verysmart!thanks!

Commented by ajfour last updated on 14/Jun/17

good question sir, couldn′t apply much geometry.

goodquestionsir,couldn′tapplymuchgeometry.

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