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Question-15784

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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.
$$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral}.\: \\ $$$$\mathrm{On}\:\mathrm{each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{it}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{constructed}. \\ $$$$\mathrm{E},\mathrm{F},\mathrm{G},\mathrm{H}\:\mathrm{are}\:\mathrm{center}\:\mathrm{points}\:\mathrm{of}\:\mathrm{these} \\ $$$$\mathrm{squares}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{EG}=\mathrm{FH}\:\mathrm{and} \\ $$$$\mathrm{EG}\bot\mathrm{FH}. \\ $$
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 .
$$\:\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} ={z}_{\mathrm{3}} +{z}_{\mathrm{4}} \\ $$$$\:\:{or}\:\:{z}_{\mathrm{1}} −{z}_{\mathrm{4}} ={z}_{\mathrm{3}} −{z}_{\mathrm{2}} \\ $$$$\:{z}_{{H}} −{z}_{{F}} =\frac{{iz}_{\mathrm{1}} }{\mathrm{2}}+\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+{z}_{\mathrm{2}} −\frac{{z}_{\mathrm{4}} }{\mathrm{2}}+\frac{{iz}_{\mathrm{4}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{i}\left({z}_{\mathrm{1}} +{z}_{\mathrm{4}} \right)}{\mathrm{2}}+\frac{\left({z}_{\mathrm{1}} −{z}_{\mathrm{4}} \right)}{\mathrm{2}}+{z}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{i}\left({z}_{\mathrm{1}} +{z}_{\mathrm{4}} \right)}{\mathrm{2}}+\frac{\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right)}{\mathrm{2}}+{z}_{\mathrm{2}} \\ $$$$\:\:{z}_{{H}} −{z}_{{F}} =\frac{{i}\left({z}_{\mathrm{1}} +{z}_{\mathrm{4}} \right)}{\mathrm{2}}+\frac{\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)}{\mathrm{2}}\:\:….\left({i}\right) \\ $$$$\:{z}_{{G}} −{z}_{{E}} =−\frac{{iz}_{\mathrm{3}} }{\mathrm{2}}+\frac{{z}_{\mathrm{3}} }{\mathrm{2}}+{z}_{\mathrm{4}} −\frac{{z}_{\mathrm{2}} }{\mathrm{2}}−\frac{{iz}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{{i}\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)}{\mathrm{2}}+\frac{\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right)}{\mathrm{2}}+{z}_{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{{i}\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)}{\mathrm{2}}+\frac{\left({z}_{\mathrm{1}} −{z}_{\mathrm{4}} \right)}{\mathrm{2}}+{z}_{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{{i}\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)}{\mathrm{2}}+\frac{\left({z}_{\mathrm{1}} +{z}_{\mathrm{4}} \right)}{\mathrm{2}}\:. \\ $$$${Now} \\ $$$$\:\left({z}_{{G}} −{z}_{{E}} \right)×{i}\:=\frac{\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)}{\mathrm{2}}+\frac{{i}\left({z}_{\mathrm{1}} +{z}_{\mathrm{4}} \right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{z}_{{H}} −{z}_{{F}} \\ $$$$\Rightarrow\:\:\mid{z}_{{G}} −{z}_{{E}} \mid\:=\:\mid{z}_{{H}} −{z}_{{F}} \mid \\ $$$$\:\:{or}\:\:\:{length}\:\:{EG}\:=\:{length}\:{FH} \\ $$$${and}\:\:{z}_{{H}} −{z}_{{F}} \:={e}^{{i}\pi/\mathrm{2}} \left({z}_{{G}} −{z}_{{E}} \right) \\ $$$${which}\:{means}\:{line}\:{FH}\:{is}\:{at}\:\mathrm{90}° \\ $$$$\:{counterclockwise}\:{to}\:{EG}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by mrW1 last updated on 14/Jun/17
very smart! thanks!
$$\mathrm{very}\:\mathrm{smart}!\:\mathrm{thanks}! \\ $$
Commented by ajfour last updated on 14/Jun/17
good question sir, couldn′t apply much geometry.
$${good}\:{question}\:{sir},\:{couldn}'{t}\:{apply} \\ $$$${much}\:{geometry}. \\ $$

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