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Question Number 15927 by ajfour last updated on 15/Jun/17

Commented by ajfour last updated on 15/Jun/17

Q.15917 (sorry it got uploaded as new question) mid points of the sides of quadrilateral ABCD are E, F, G, and H. Midpoints of its digonals L and K. To prove : intersection point J of EG and HF lies on LK and is their (EG, HF, LK) midpoint.

$${Q}.\mathrm{15917}\:\left({sorry}\:{it}\:{got}\:{uploaded}\right. \\ $$$$\left.{as}\:{new}\:{question}\right) \\ $$$${mid}\:{points}\:{of}\:{the}\:{sides}\:{of}\: \\ $$$${quadrilateral}\:{ABCD}\:{are}\: \\ $$$${E},\:{F},\:{G},\:{and}\:{H}.\:{Midpoints}\:{of} \\ $$$${its}\:{digonals}\:{L}\:{and}\:{K}. \\ $$$$\:\:{To}\:{prove}\::\:{intersection}\:{point}\:{J} \\ $$$${of}\:{EG}\:{and}\:{HF}\:\:\:{lies}\:{on}\:{LK}\:{and} \\ $$$${is}\:{their}\:\left({EG},\:{HF},\:{LK}\right)\:{midpoint}. \\ $$

Commented by ajfour last updated on 15/Jun/17

z_1 +z_2 =z_3 +z_4 .....(i) let B is the origin z_L =((z_1 +z_2 )/2) , z_K =z_1 +((z_3 −z_1 )/2) =((z_1 +z_3 )/2) z_H =(z_1 /2) , z_F =z_3 +(z_4 /2) A point of HF : z_(HF) = z_H +μ(z_F −z_H ) = (z_1 /2)+μ(z_3 +((z_4 −z_1 )/2)) = (z_1 /2)+μ(z_3 +((z_2 −z_3 )/2)) [see (i)] z_(HF) = (z_1 /2)+μ(((z_2 +z_3 )/2)) ...(ii) z_E = (z_3 /2) , z_G = z_1 +(z_2 /2) A point on EG obeys : z_(EG) = z_E +ε(z_G −z_E ) = (z_3 /2)+ε(z_1 +((z_2 −z_3 )/2)) ...(iii) point of intersection J of HF and EG shall obey both their eqns. (ii) and (iii) z_J =(z_1 /2)+μ(((z_2 +z_3 )/2))=(z_3 /2)+ε(z_1 +((z_2 −z_3 )/2)) ⇒ z_1 ((1/2)−ε)+z_2 ((μ/2)−(ε/2))+ z_3 ((μ/2)−(1/2)+(ε/2)) =0 as z_1 , z_2 , z_3 can be independently chosen to form a quadrilateral but above eqn. is always true, so for point J, ε=(1/2) , μ=ε=(1/2) this implies J is the midpoint of HF and EG , and that z_J =(z_1 /2)+((z_2 +z_3 )/4) (by substituting μ=(1/2) in (ii)) It now remains to prove that J is also the midpoint of LK. midpoint of LK is z=((z_L +z_K )/2) z= (1/2)(((z_1 +z_2 )/2)+((z_1 +z_3 )/2)) = (z_1 /2)+((z_2 +z_3 )/4) = z_(J ) so midpoints of EG, HF, and LK coincide in J .

$$\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} ={z}_{\mathrm{3}} +{z}_{\mathrm{4}} \:\:\:.....\left({i}\right) \\ $$$${let}\:{B}\:{is}\:{the}\:{origin}\: \\ $$$$\:{z}_{{L}} =\frac{{z}_{\mathrm{1}} +{z}_{\mathrm{2}} }{\mathrm{2}}\:\:, \\ $$$$\:{z}_{{K}} ={z}_{\mathrm{1}} +\frac{{z}_{\mathrm{3}} −{z}_{\mathrm{1}} }{\mathrm{2}}\:=\frac{{z}_{\mathrm{1}} +{z}_{\mathrm{3}} }{\mathrm{2}} \\ $$$$\:{z}_{{H}} =\frac{{z}_{\mathrm{1}} }{\mathrm{2}}\:\:,\:\:{z}_{{F}} ={z}_{\mathrm{3}} +\frac{{z}_{\mathrm{4}} }{\mathrm{2}}\: \\ $$$$\:{A}\:{point}\:{of}\:{HF}\:: \\ $$$$\:\:{z}_{{HF}} \:=\:{z}_{{H}} +\mu\left({z}_{{F}} −{z}_{{H}} \right) \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+\mu\left({z}_{\mathrm{3}} +\frac{{z}_{\mathrm{4}} −{z}_{\mathrm{1}} }{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+\mu\left({z}_{\mathrm{3}} +\frac{{z}_{\mathrm{2}} −{z}_{\mathrm{3}} }{\mathrm{2}}\right)\:\:\:\:\left[{see}\:\left({i}\right)\right] \\ $$$$\:\:\:\:\:{z}_{{HF}} \:=\:\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+\mu\left(\frac{{z}_{\mathrm{2}} +{z}_{\mathrm{3}} }{\mathrm{2}}\right)\:\:\:\:...\left({ii}\right) \\ $$$$\:\:{z}_{{E}} =\:\frac{{z}_{\mathrm{3}} }{\mathrm{2}}\:\:,\:\:{z}_{{G}} \:=\:{z}_{\mathrm{1}} +\frac{{z}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:{A}\:{point}\:{on}\:{EG}\:{obeys}\:: \\ $$$$\:{z}_{{EG}} =\:{z}_{{E}} +\epsilon\left({z}_{{G}} −{z}_{{E}} \right) \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{z}_{\mathrm{3}} }{\mathrm{2}}+\epsilon\left({z}_{\mathrm{1}} +\frac{{z}_{\mathrm{2}} −{z}_{\mathrm{3}} }{\mathrm{2}}\right)\:\:\:\:...\left({iii}\right) \\ $$$${point}\:{of}\:{intersection}\:{J}\:{of}\:{HF}\:{and} \\ $$$${EG}\:{shall}\:{obey}\:{both}\:{their}\:{eqns}. \\ $$$$\:\:\left({ii}\right)\:{and}\:\left({iii}\right) \\ $$$${z}_{{J}} =\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+\mu\left(\frac{{z}_{\mathrm{2}} +{z}_{\mathrm{3}} }{\mathrm{2}}\right)=\frac{{z}_{\mathrm{3}} }{\mathrm{2}}+\epsilon\left({z}_{\mathrm{1}} +\frac{{z}_{\mathrm{2}} −{z}_{\mathrm{3}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:{z}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\epsilon\right)+{z}_{\mathrm{2}} \left(\frac{\mu}{\mathrm{2}}−\frac{\epsilon}{\mathrm{2}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}_{\mathrm{3}} \left(\frac{\mu}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\epsilon}{\mathrm{2}}\right)\:=\mathrm{0} \\ $$$$\:\:{as}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:{can}\:{be}\:{independently} \\ $$$$\:{chosen}\:{to}\:{form}\:{a}\:{quadrilateral} \\ $$$$\:{but}\:{above}\:{eqn}.\:{is}\:{always}\:{true},\:{so} \\ $$$${for}\:{point}\:{J}, \\ $$$$\:\:\:\epsilon=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mu=\epsilon=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${this}\:{implies}\:{J}\:{is}\:{the}\:{midpoint} \\ $$$${of}\:{HF}\:{and}\:{EG}\:,\:{and}\:{that} \\ $$$$\:{z}_{{J}} =\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+\frac{{z}_{\mathrm{2}} +{z}_{\mathrm{3}} }{\mathrm{4}}\:\:\: \\ $$$$\:\:\left({by}\:{substituting}\:\mu=\frac{\mathrm{1}}{\mathrm{2}}\:{in}\:\left({ii}\right)\right) \\ $$$${It}\:{now}\:{remains}\:{to}\:{prove}\:{that} \\ $$$${J}\:{is}\:{also}\:{the}\:{midpoint}\:{of}\:{LK}. \\ $$$$\:{midpoint}\:{of}\:{LK}\:{is} \\ $$$$\:\:\:\:\:{z}=\frac{{z}_{{L}} +{z}_{{K}} }{\mathrm{2}} \\ $$$$\:\:\:\:{z}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{z}_{\mathrm{1}} +{z}_{\mathrm{2}} }{\mathrm{2}}+\frac{{z}_{\mathrm{1}} +{z}_{\mathrm{3}} }{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:=\:\frac{{z}_{\mathrm{1}} }{\mathrm{2}}+\frac{{z}_{\mathrm{2}} +{z}_{\mathrm{3}} }{\mathrm{4}}\:\:\:=\:{z}_{{J}\:} \\ $$$${so}\:{midpoints}\:{of}\:{EG},\:{HF},\:{and} \\ $$$$\:{LK}\:{coincide}\:{in}\:{J}\:. \\ $$

Commented by mrW1 last updated on 15/Jun/17

excellent sir! you master the vector technique outstandingly.

$$\mathrm{excellent}\:\mathrm{sir}!\:\mathrm{you}\:\mathrm{master}\:\mathrm{the}\:\mathrm{vector} \\ $$$$\mathrm{technique}\:\mathrm{outstandingly}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17

let:A(a),B(b),C(c),D(d),L(l),K(k) H(h)=((b+c)/2),F(f)=((a+d)/2),E(e)=((a+b)/2), G(g)=((d+c)/2) ⇒L(l)=((b+d)/2),K(k)=((a+c)/2) J=((((b+d)/2)+((a+c)/2))/2)=((a+b+c+d)/4) (i) midpoint of FH=((((a+d)/2)+((b+c)/2))/2)=((a+d+b+c)/4) (ii) midpoint of EG=((((a+b)/2)+((d+c)/2))/2)=((a+b+d+c)/4) (iii) a,b,c,d,e,f,g,,k,l,are complex numbers. from equations i,ii,iii,we see that: midpoint of:EG,FH,LK,have the same cordinates.so they should be the same point. ■

$${let}:{A}\left({a}\right),{B}\left({b}\right),{C}\left({c}\right),{D}\left({d}\right),{L}\left({l}\right),{K}\left({k}\right) \\ $$$${H}\left({h}\right)=\frac{{b}+{c}}{\mathrm{2}},{F}\left({f}\right)=\frac{{a}+{d}}{\mathrm{2}},{E}\left({e}\right)=\frac{{a}+{b}}{\mathrm{2}}, \\ $$$${G}\left({g}\right)=\frac{{d}+{c}}{\mathrm{2}} \\ $$$$\Rightarrow{L}\left({l}\right)=\frac{{b}+{d}}{\mathrm{2}},{K}\left({k}\right)=\frac{{a}+{c}}{\mathrm{2}} \\ $$$${J}=\frac{\frac{{b}+{d}}{\mathrm{2}}+\frac{{a}+{c}}{\mathrm{2}}}{\mathrm{2}}=\frac{{a}+{b}+{c}+{d}}{\mathrm{4}}\:\:\left({i}\right) \\ $$$${midpoint}\:{of}\:{FH}=\frac{\frac{{a}+{d}}{\mathrm{2}}+\frac{{b}+{c}}{\mathrm{2}}}{\mathrm{2}}=\frac{{a}+{d}+{b}+{c}}{\mathrm{4}}\:\left({ii}\right) \\ $$$$ \\ $$$${midpoint}\:{of}\:{EG}=\frac{\frac{{a}+{b}}{\mathrm{2}}+\frac{{d}+{c}}{\mathrm{2}}}{\mathrm{2}}=\frac{{a}+{b}+{d}+{c}}{\mathrm{4}}\:\left({iii}\right) \\ $$$${a},{b},{c},{d},{e},{f},{g},,{k},{l},{are}\:{complex}\:{numbers}. \\ $$$${from}\:{equations}\:{i},{ii},{iii},{we}\:{see}\:{that}: \\ $$$${midpoint}\:{of}:{EG},{FH},{LK},{have}\:{the}\:{same} \\ $$$${cordinates}.{so}\:{they}\:{should}\:{be}\:{the}\:{same} \\ $$$${point}.\:\blacksquare \\ $$

Commented by ajfour last updated on 19/Jun/17

thank you, sir. you cut my long story short.

$${thank}\:{you},\:{sir}.\:{you}\:{cut}\:{my}\:{long} \\ $$$${story}\:{short}. \\ $$

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