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Question Number 15917 by mrW1 last updated on 15/Jun/17

Commented by mrW1 last updated on 15/Jun/17

E,F,G,H are mid−points of the sides of a convex quadrilateral ABCD. L,K are mid−points of the diagonals. Prove that EG,FH,LK intersect at the same point J and J is the mid−point of them.

$$\mathrm{E},\mathrm{F},\mathrm{G},\mathrm{H}\:\mathrm{are}\:\mathrm{mid}−\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral}\:\mathrm{ABCD}. \\ $$$$\mathrm{L},\mathrm{K}\:\mathrm{are}\:\mathrm{mid}−\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagonals}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{EG},\mathrm{FH},\mathrm{LK}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{point}\:\mathrm{J}\:\:\mathrm{and}\:\mathrm{J}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\:\mathrm{them}. \\ $$

Answered by mrW1 last updated on 16/Jun/17

An other way to solve: FK=(1/2)CD LH=(1/2)CD ⇒FK=LH and FK∥LH FL=(1/2)AB HK=(1/2)AB ⇒FL=HK and FL∥HK ⇒HLFK is parallelogram ⇒LJ=JK and FJ=JH similarly EL=(1/2)AD KG=(1/2)AD ⇒EL=KG and EL∥KG EK=(1/2)BC LG=(1/2)BC ⇒EK=LG and EK∥LG ⇒ELGK is parallelogram ⇒LJ=JK and EJ=JG ⇒J is mid−point of LK,FH and EG.

$$\mathrm{An}\:\mathrm{other}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}: \\ $$$$\mathrm{FK}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{CD} \\ $$$$\mathrm{LH}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{CD} \\ $$$$\Rightarrow\mathrm{FK}=\mathrm{LH}\:\mathrm{and}\:\mathrm{FK}\parallel\mathrm{LH} \\ $$$$ \\ $$$$\mathrm{FL}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB} \\ $$$$\mathrm{HK}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB} \\ $$$$\Rightarrow\mathrm{FL}=\mathrm{HK}\:\mathrm{and}\:\mathrm{FL}\parallel\mathrm{HK} \\ $$$$ \\ $$$$\Rightarrow\mathrm{HLFK}\:\mathrm{is}\:\mathrm{parallelogram} \\ $$$$\Rightarrow\mathrm{LJ}=\mathrm{JK}\:\mathrm{and}\:\mathrm{FJ}=\mathrm{JH} \\ $$$$ \\ $$$$\mathrm{similarly} \\ $$$$\mathrm{EL}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AD} \\ $$$$\mathrm{KG}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AD} \\ $$$$\Rightarrow\mathrm{EL}=\mathrm{KG}\:\mathrm{and}\:\mathrm{EL}\parallel\mathrm{KG} \\ $$$$ \\ $$$$\mathrm{EK}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{BC} \\ $$$$\mathrm{LG}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{BC} \\ $$$$\Rightarrow\mathrm{EK}=\mathrm{LG}\:\mathrm{and}\:\mathrm{EK}\parallel\mathrm{LG} \\ $$$$ \\ $$$$\Rightarrow\mathrm{ELGK}\:\mathrm{is}\:\mathrm{parallelogram} \\ $$$$\Rightarrow\mathrm{LJ}=\mathrm{JK}\:\mathrm{and}\:\mathrm{EJ}=\mathrm{JG} \\ $$$$ \\ $$$$\Rightarrow\mathrm{J}\:\mathrm{is}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\mathrm{LK},\mathrm{FH}\:\mathrm{and}\:\mathrm{EG}. \\ $$

Commented by mrW1 last updated on 16/Jun/17

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