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

Commented by mrW1 last updated on 15/Jun/17

$E, F, G, H are mid - points of the sides$ $of a convex quadrilateral ABCD .$ $L, K are mid - points of the diagonals .$ $Prove that EG, FH, LK intersect at the$ $same point J and J is the mid - point of them .$
	Answered by mrW1 last updated on 16/Jun/17

	$An other way to solve :$ $FK = \frac{1}{2} CD$ $LH = \frac{1}{2} CD$ $\Rightarrow FK = LH and FK ∥ LH$ $FL = \frac{1}{2} AB$ $HK = \frac{1}{2} AB$ $\Rightarrow FL = HK and FL ∥ HK$ $\Rightarrow HLFK is parallelogram$ $\Rightarrow LJ = JK and FJ = JH$ $similarly$ $EL = \frac{1}{2} AD$ $KG = \frac{1}{2} AD$ $\Rightarrow EL = KG and EL ∥ KG$ $EK = \frac{1}{2} BC$ $LG = \frac{1}{2} BC$ $\Rightarrow EK = LG and EK ∥ LG$ $\Rightarrow ELGK is parallelogram$ $\Rightarrow LJ = JK and EJ = JG$ $\Rightarrow J is mid - point of LK, FH and EG .$
	Commented by mrW1 last updated on 16/Jun/17

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