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Euler Line in a quadrilateral
Entrepreneur   1
N Yesterday at 8:23 AM by Entrepreneur
Let $ABCD$ be a convex quadrilateral and let $G_a,O_a,H_a\;\&\;N_a$ be the centroid, circumcentre, orthocenter & nine-point centre of $\Delta BCD$ respectively. We define the points $G_b,G_c,G_d,$ $O_b,O_c,O_d,$ $H_b,H_c,H_d$ $N_b,N_c\;\&\;N_d$ analogously. Now, we define the area centroid $\cal G$ of $ABCD$ as the intersection of $G_a,G_c\;\&\;G_bG_d,$ the qausicircumcentre $\cal O$ as the intersection of $ O_aO_c\;\&\;O_bO_d,$ quasiorthocentre $\cal H$ as the intersection of $H_aH_c\;\&\;H_bH_d$ and the quasinine-point centre $\cal N$ as the intersection of $N_aN_c\;\&\;N_bN_d.$ Prove that the points $\mathcal{G,O,H,N}$ are collinear with $\mathcal{HG}=2\mathcal{GO}$ and $\mathcal{HN}=\mathcal{NO}.$
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Entrepreneur
Nov 9, 2024
Entrepreneur
Yesterday at 8:23 AM
Euler Line in a quadrilateral
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Let $ABCD$ be a convex quadrilateral and let $G_a,O_a,H_a\;\&\;N_a$ be the centroid, circumcentre, orthocenter & nine-point centre of $\Delta BCD$ respectively. We define the points $G_b,G_c,G_d,$ $O_b,O_c,O_d,$ $H_b,H_c,H_d$ $N_b,N_c\;\&\;N_d$ analogously. Now, we define the area centroid $\cal G$ of $ABCD$ as the intersection of $G_a,G_c\;\&\;G_bG_d,$ the qausicircumcentre $\cal O$ as the intersection of $ O_aO_c\;\&\;O_bO_d,$ quasiorthocentre $\cal H$ as the intersection of $H_aH_c\;\&\;H_bH_d$ and the quasinine-point centre $\cal N$ as the intersection of $N_aN_c\;\&\;N_bN_d.$ Prove that the points $\mathcal{G,O,H,N}$ are collinear with $\mathcal{HG}=2\mathcal{GO}$ and $\mathcal{HN}=\mathcal{NO}.$
This post has been edited 1 time. Last edited by Entrepreneur, Yesterday at 8:23 AM
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Bump....
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