Rhombus in a cyclic quadrilateral











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Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. The lines $AB$ and $CD$ intersect at point $P$. The lines $AD$ and $BC$ intersect in point $Q$. The bisector of the angle $angle DPA$ cuts the line segment $BC$ and $DA$ in the points $E$ and $G$, respectively. The bisector of the angle $angle AQB$ cuts the line segments $AB$ and $CD$ in the points $H$ and $F$.



Now it seems as if the quadrilateral $EFGH$ is a always a rhombus. I intend to prove this.



This is a sketch of the problem



Maybe anyone has a checklist or any idea to begin with.










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    @Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic?
    – Narasimham
    Nov 17 at 21:03












  • Yes, this is why I've constructed the circle
    – calculatormathematical
    Nov 17 at 23:53















up vote
3
down vote

favorite
1












Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. The lines $AB$ and $CD$ intersect at point $P$. The lines $AD$ and $BC$ intersect in point $Q$. The bisector of the angle $angle DPA$ cuts the line segment $BC$ and $DA$ in the points $E$ and $G$, respectively. The bisector of the angle $angle AQB$ cuts the line segments $AB$ and $CD$ in the points $H$ and $F$.



Now it seems as if the quadrilateral $EFGH$ is a always a rhombus. I intend to prove this.



This is a sketch of the problem



Maybe anyone has a checklist or any idea to begin with.










share|cite|improve this question




















  • 2




    @Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic?
    – Narasimham
    Nov 17 at 21:03












  • Yes, this is why I've constructed the circle
    – calculatormathematical
    Nov 17 at 23:53













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. The lines $AB$ and $CD$ intersect at point $P$. The lines $AD$ and $BC$ intersect in point $Q$. The bisector of the angle $angle DPA$ cuts the line segment $BC$ and $DA$ in the points $E$ and $G$, respectively. The bisector of the angle $angle AQB$ cuts the line segments $AB$ and $CD$ in the points $H$ and $F$.



Now it seems as if the quadrilateral $EFGH$ is a always a rhombus. I intend to prove this.



This is a sketch of the problem



Maybe anyone has a checklist or any idea to begin with.










share|cite|improve this question















Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. The lines $AB$ and $CD$ intersect at point $P$. The lines $AD$ and $BC$ intersect in point $Q$. The bisector of the angle $angle DPA$ cuts the line segment $BC$ and $DA$ in the points $E$ and $G$, respectively. The bisector of the angle $angle AQB$ cuts the line segments $AB$ and $CD$ in the points $H$ and $F$.



Now it seems as if the quadrilateral $EFGH$ is a always a rhombus. I intend to prove this.



This is a sketch of the problem



Maybe anyone has a checklist or any idea to begin with.







geometry proof-writing euclidean-geometry circle quadrilateral






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share|cite|improve this question













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edited Nov 17 at 21:17









Batominovski

32.3k23190




32.3k23190










asked Nov 17 at 19:06









calculatormathematical

3811




3811








  • 2




    @Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic?
    – Narasimham
    Nov 17 at 21:03












  • Yes, this is why I've constructed the circle
    – calculatormathematical
    Nov 17 at 23:53














  • 2




    @Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic?
    – Narasimham
    Nov 17 at 21:03












  • Yes, this is why I've constructed the circle
    – calculatormathematical
    Nov 17 at 23:53








2




2




@Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic?
– Narasimham
Nov 17 at 21:03






@Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic?
– Narasimham
Nov 17 at 21:03














Yes, this is why I've constructed the circle
– calculatormathematical
Nov 17 at 23:53




Yes, this is why I've constructed the circle
– calculatormathematical
Nov 17 at 23:53










2 Answers
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Here is an alternative way to show that $EGperp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).



Let $EG$ and $FH$ meet at $S$. Write $alpha$, $beta$, $gamma$, and $delta$ for the angles $angle DAB$, $angle ABC$, $angle BCD$, and $angle CDA$, respectively. Then,
$$angle CQD=pi-gamma-deltatext{ so }angle SQB=frac{angle CQD}{2}=frac{pi}{2}-frac{gamma}{2}-frac{delta}{2},.$$
Similarly,
$$angle AQD=pi-alpha-deltatext{ so }angle SPB=frac{angle CQD}{2}=frac{pi}{2}-frac{alpha}{2}-frac{delta}{2},.$$
The sum of internal angles of the quadrilateral $PSQB$ is $2pi$, whence
$$begin{align}angle PSQ&=2pi-big(angle SQB+angle SPB+(2pi-angle PBQ)big)\&=angle PBQ - angle SQB-angle SPB,.end{align}$$
However, $angle PBQ=angle ABC=beta$, so
$$angle PSQ=beta-left(frac{pi}{2}-frac{gamma}{2}-frac{delta}{2}right)-left(frac{pi}{2}-frac{alpha}{2}-frac{delta}{2}right),.$$
That is,
$$angle PSQ=(beta+delta)+frac{alpha+gamma}{2}-pi=frac{beta+delta}{2}+frac{alpha+beta+gamma+delta}{2}-pi,.$$
Since $alpha+beta+gamma+delta=2pi$, we have
$$angle PSQ=frac{beta+delta}{2},.$$



If $EGHF$ is a rhombus, then $angle PSQ=dfrac{pi}{2}$, making $beta+delta=pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $beta+delta=pi$ implies that $angle PSQ=dfrac{pi}{2}$, so $EGperp HF$. The rest goes as Marco's answer.






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    We first show that $EG perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have
    $$begin{align}newcommand{arc}[1]{overset{mmlToken{mo}{⏜}}{#1}}arc{H'E'}+arc{G'F'}&=arc{H'B}+arc{BE'}+arc{G'D}+arc{DF'}=(arc{H'B}+arc{DF'})+(arc{BE'}+arc{G'D})\&=(arc{CF'}+arc{AH'})+(arc{E'C}+arc{AG'})=arc{E'C}+arc{CF'}+arc{H'A}+arc{AG'}\&=arc{E'F'}+arc{H'G'},end{align}$$
    which implies our claim.



    Let $S$ be the intersection of $EG$ and $HF$.
    Now, in triangle $triangle PHF$ the angle bisector of $angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.






    share|cite|improve this answer























    • I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
      – Batominovski
      Nov 17 at 22:05












    • @Batominovski looks good thank you.
      – Marco
      Nov 18 at 2:49











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    Here is an alternative way to show that $EGperp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).



    Let $EG$ and $FH$ meet at $S$. Write $alpha$, $beta$, $gamma$, and $delta$ for the angles $angle DAB$, $angle ABC$, $angle BCD$, and $angle CDA$, respectively. Then,
    $$angle CQD=pi-gamma-deltatext{ so }angle SQB=frac{angle CQD}{2}=frac{pi}{2}-frac{gamma}{2}-frac{delta}{2},.$$
    Similarly,
    $$angle AQD=pi-alpha-deltatext{ so }angle SPB=frac{angle CQD}{2}=frac{pi}{2}-frac{alpha}{2}-frac{delta}{2},.$$
    The sum of internal angles of the quadrilateral $PSQB$ is $2pi$, whence
    $$begin{align}angle PSQ&=2pi-big(angle SQB+angle SPB+(2pi-angle PBQ)big)\&=angle PBQ - angle SQB-angle SPB,.end{align}$$
    However, $angle PBQ=angle ABC=beta$, so
    $$angle PSQ=beta-left(frac{pi}{2}-frac{gamma}{2}-frac{delta}{2}right)-left(frac{pi}{2}-frac{alpha}{2}-frac{delta}{2}right),.$$
    That is,
    $$angle PSQ=(beta+delta)+frac{alpha+gamma}{2}-pi=frac{beta+delta}{2}+frac{alpha+beta+gamma+delta}{2}-pi,.$$
    Since $alpha+beta+gamma+delta=2pi$, we have
    $$angle PSQ=frac{beta+delta}{2},.$$



    If $EGHF$ is a rhombus, then $angle PSQ=dfrac{pi}{2}$, making $beta+delta=pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $beta+delta=pi$ implies that $angle PSQ=dfrac{pi}{2}$, so $EGperp HF$. The rest goes as Marco's answer.






    share|cite|improve this answer



























      up vote
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      Here is an alternative way to show that $EGperp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).



      Let $EG$ and $FH$ meet at $S$. Write $alpha$, $beta$, $gamma$, and $delta$ for the angles $angle DAB$, $angle ABC$, $angle BCD$, and $angle CDA$, respectively. Then,
      $$angle CQD=pi-gamma-deltatext{ so }angle SQB=frac{angle CQD}{2}=frac{pi}{2}-frac{gamma}{2}-frac{delta}{2},.$$
      Similarly,
      $$angle AQD=pi-alpha-deltatext{ so }angle SPB=frac{angle CQD}{2}=frac{pi}{2}-frac{alpha}{2}-frac{delta}{2},.$$
      The sum of internal angles of the quadrilateral $PSQB$ is $2pi$, whence
      $$begin{align}angle PSQ&=2pi-big(angle SQB+angle SPB+(2pi-angle PBQ)big)\&=angle PBQ - angle SQB-angle SPB,.end{align}$$
      However, $angle PBQ=angle ABC=beta$, so
      $$angle PSQ=beta-left(frac{pi}{2}-frac{gamma}{2}-frac{delta}{2}right)-left(frac{pi}{2}-frac{alpha}{2}-frac{delta}{2}right),.$$
      That is,
      $$angle PSQ=(beta+delta)+frac{alpha+gamma}{2}-pi=frac{beta+delta}{2}+frac{alpha+beta+gamma+delta}{2}-pi,.$$
      Since $alpha+beta+gamma+delta=2pi$, we have
      $$angle PSQ=frac{beta+delta}{2},.$$



      If $EGHF$ is a rhombus, then $angle PSQ=dfrac{pi}{2}$, making $beta+delta=pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $beta+delta=pi$ implies that $angle PSQ=dfrac{pi}{2}$, so $EGperp HF$. The rest goes as Marco's answer.






      share|cite|improve this answer

























        up vote
        2
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        up vote
        2
        down vote









        Here is an alternative way to show that $EGperp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).



        Let $EG$ and $FH$ meet at $S$. Write $alpha$, $beta$, $gamma$, and $delta$ for the angles $angle DAB$, $angle ABC$, $angle BCD$, and $angle CDA$, respectively. Then,
        $$angle CQD=pi-gamma-deltatext{ so }angle SQB=frac{angle CQD}{2}=frac{pi}{2}-frac{gamma}{2}-frac{delta}{2},.$$
        Similarly,
        $$angle AQD=pi-alpha-deltatext{ so }angle SPB=frac{angle CQD}{2}=frac{pi}{2}-frac{alpha}{2}-frac{delta}{2},.$$
        The sum of internal angles of the quadrilateral $PSQB$ is $2pi$, whence
        $$begin{align}angle PSQ&=2pi-big(angle SQB+angle SPB+(2pi-angle PBQ)big)\&=angle PBQ - angle SQB-angle SPB,.end{align}$$
        However, $angle PBQ=angle ABC=beta$, so
        $$angle PSQ=beta-left(frac{pi}{2}-frac{gamma}{2}-frac{delta}{2}right)-left(frac{pi}{2}-frac{alpha}{2}-frac{delta}{2}right),.$$
        That is,
        $$angle PSQ=(beta+delta)+frac{alpha+gamma}{2}-pi=frac{beta+delta}{2}+frac{alpha+beta+gamma+delta}{2}-pi,.$$
        Since $alpha+beta+gamma+delta=2pi$, we have
        $$angle PSQ=frac{beta+delta}{2},.$$



        If $EGHF$ is a rhombus, then $angle PSQ=dfrac{pi}{2}$, making $beta+delta=pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $beta+delta=pi$ implies that $angle PSQ=dfrac{pi}{2}$, so $EGperp HF$. The rest goes as Marco's answer.






        share|cite|improve this answer














        Here is an alternative way to show that $EGperp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).



        Let $EG$ and $FH$ meet at $S$. Write $alpha$, $beta$, $gamma$, and $delta$ for the angles $angle DAB$, $angle ABC$, $angle BCD$, and $angle CDA$, respectively. Then,
        $$angle CQD=pi-gamma-deltatext{ so }angle SQB=frac{angle CQD}{2}=frac{pi}{2}-frac{gamma}{2}-frac{delta}{2},.$$
        Similarly,
        $$angle AQD=pi-alpha-deltatext{ so }angle SPB=frac{angle CQD}{2}=frac{pi}{2}-frac{alpha}{2}-frac{delta}{2},.$$
        The sum of internal angles of the quadrilateral $PSQB$ is $2pi$, whence
        $$begin{align}angle PSQ&=2pi-big(angle SQB+angle SPB+(2pi-angle PBQ)big)\&=angle PBQ - angle SQB-angle SPB,.end{align}$$
        However, $angle PBQ=angle ABC=beta$, so
        $$angle PSQ=beta-left(frac{pi}{2}-frac{gamma}{2}-frac{delta}{2}right)-left(frac{pi}{2}-frac{alpha}{2}-frac{delta}{2}right),.$$
        That is,
        $$angle PSQ=(beta+delta)+frac{alpha+gamma}{2}-pi=frac{beta+delta}{2}+frac{alpha+beta+gamma+delta}{2}-pi,.$$
        Since $alpha+beta+gamma+delta=2pi$, we have
        $$angle PSQ=frac{beta+delta}{2},.$$



        If $EGHF$ is a rhombus, then $angle PSQ=dfrac{pi}{2}$, making $beta+delta=pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $beta+delta=pi$ implies that $angle PSQ=dfrac{pi}{2}$, so $EGperp HF$. The rest goes as Marco's answer.







        share|cite|improve this answer














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        share|cite|improve this answer








        edited Nov 17 at 21:54

























        answered Nov 17 at 21:39









        Batominovski

        32.3k23190




        32.3k23190






















            up vote
            2
            down vote













            We first show that $EG perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have
            $$begin{align}newcommand{arc}[1]{overset{mmlToken{mo}{⏜}}{#1}}arc{H'E'}+arc{G'F'}&=arc{H'B}+arc{BE'}+arc{G'D}+arc{DF'}=(arc{H'B}+arc{DF'})+(arc{BE'}+arc{G'D})\&=(arc{CF'}+arc{AH'})+(arc{E'C}+arc{AG'})=arc{E'C}+arc{CF'}+arc{H'A}+arc{AG'}\&=arc{E'F'}+arc{H'G'},end{align}$$
            which implies our claim.



            Let $S$ be the intersection of $EG$ and $HF$.
            Now, in triangle $triangle PHF$ the angle bisector of $angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.






            share|cite|improve this answer























            • I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
              – Batominovski
              Nov 17 at 22:05












            • @Batominovski looks good thank you.
              – Marco
              Nov 18 at 2:49















            up vote
            2
            down vote













            We first show that $EG perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have
            $$begin{align}newcommand{arc}[1]{overset{mmlToken{mo}{⏜}}{#1}}arc{H'E'}+arc{G'F'}&=arc{H'B}+arc{BE'}+arc{G'D}+arc{DF'}=(arc{H'B}+arc{DF'})+(arc{BE'}+arc{G'D})\&=(arc{CF'}+arc{AH'})+(arc{E'C}+arc{AG'})=arc{E'C}+arc{CF'}+arc{H'A}+arc{AG'}\&=arc{E'F'}+arc{H'G'},end{align}$$
            which implies our claim.



            Let $S$ be the intersection of $EG$ and $HF$.
            Now, in triangle $triangle PHF$ the angle bisector of $angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.






            share|cite|improve this answer























            • I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
              – Batominovski
              Nov 17 at 22:05












            • @Batominovski looks good thank you.
              – Marco
              Nov 18 at 2:49













            up vote
            2
            down vote










            up vote
            2
            down vote









            We first show that $EG perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have
            $$begin{align}newcommand{arc}[1]{overset{mmlToken{mo}{⏜}}{#1}}arc{H'E'}+arc{G'F'}&=arc{H'B}+arc{BE'}+arc{G'D}+arc{DF'}=(arc{H'B}+arc{DF'})+(arc{BE'}+arc{G'D})\&=(arc{CF'}+arc{AH'})+(arc{E'C}+arc{AG'})=arc{E'C}+arc{CF'}+arc{H'A}+arc{AG'}\&=arc{E'F'}+arc{H'G'},end{align}$$
            which implies our claim.



            Let $S$ be the intersection of $EG$ and $HF$.
            Now, in triangle $triangle PHF$ the angle bisector of $angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.






            share|cite|improve this answer














            We first show that $EG perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have
            $$begin{align}newcommand{arc}[1]{overset{mmlToken{mo}{⏜}}{#1}}arc{H'E'}+arc{G'F'}&=arc{H'B}+arc{BE'}+arc{G'D}+arc{DF'}=(arc{H'B}+arc{DF'})+(arc{BE'}+arc{G'D})\&=(arc{CF'}+arc{AH'})+(arc{E'C}+arc{AG'})=arc{E'C}+arc{CF'}+arc{H'A}+arc{AG'}\&=arc{E'F'}+arc{H'G'},end{align}$$
            which implies our claim.



            Let $S$ be the intersection of $EG$ and $HF$.
            Now, in triangle $triangle PHF$ the angle bisector of $angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 17 at 22:05









            Batominovski

            32.3k23190




            32.3k23190










            answered Nov 17 at 20:47









            Marco

            2,074110




            2,074110












            • I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
              – Batominovski
              Nov 17 at 22:05












            • @Batominovski looks good thank you.
              – Marco
              Nov 18 at 2:49


















            • I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
              – Batominovski
              Nov 17 at 22:05












            • @Batominovski looks good thank you.
              – Marco
              Nov 18 at 2:49
















            I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
            – Batominovski
            Nov 17 at 22:05






            I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking.
            – Batominovski
            Nov 17 at 22:05














            @Batominovski looks good thank you.
            – Marco
            Nov 18 at 2:49




            @Batominovski looks good thank you.
            – Marco
            Nov 18 at 2:49


















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