Two tangent circles and parallel lines












1












$begingroup$


I have problems in solving the following problem:



Consider two circles which have only one point $A$ in common, i.e. which are tangent to each other. Now consider two lines through A, such that the lines meet the circles at further points $B,C,D,E$. I want to prove that the lines $DE$ and $BC$ are parallel lines ($D,E$ being points on one circle and $B,C$ on the other).



I tried to use theorems like the inscribed angle theorem, but I was not succesfull so far. Does someone know how to solve this problem?



If it is possible, I only want to use geometric arguments, and not analytical arguments. Does this result have any name?



Best wishes






geometry







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$endgroup$












  • $begingroup$
    This theorem is true if the circles are of the same size.
    $endgroup$
    – Jack
    Jun 2 '15 at 6:51
















1












$begingroup$


I have problems in solving the following problem:



Consider two circles which have only one point $A$ in common, i.e. which are tangent to each other. Now consider two lines through A, such that the lines meet the circles at further points $B,C,D,E$. I want to prove that the lines $DE$ and $BC$ are parallel lines ($D,E$ being points on one circle and $B,C$ on the other).



I tried to use theorems like the inscribed angle theorem, but I was not succesfull so far. Does someone know how to solve this problem?



If it is possible, I only want to use geometric arguments, and not analytical arguments. Does this result have any name?



Best wishes






geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    This theorem is true if the circles are of the same size.
    $endgroup$
    – Jack
    Jun 2 '15 at 6:51














1












1








1


1



$begingroup$


I have problems in solving the following problem:



Consider two circles which have only one point $A$ in common, i.e. which are tangent to each other. Now consider two lines through A, such that the lines meet the circles at further points $B,C,D,E$. I want to prove that the lines $DE$ and $BC$ are parallel lines ($D,E$ being points on one circle and $B,C$ on the other).



I tried to use theorems like the inscribed angle theorem, but I was not succesfull so far. Does someone know how to solve this problem?



If it is possible, I only want to use geometric arguments, and not analytical arguments. Does this result have any name?



Best wishes






geometry







share|cite|improve this question









$endgroup$




I have problems in solving the following problem:



Consider two circles which have only one point $A$ in common, i.e. which are tangent to each other. Now consider two lines through A, such that the lines meet the circles at further points $B,C,D,E$. I want to prove that the lines $DE$ and $BC$ are parallel lines ($D,E$ being points on one circle and $B,C$ on the other).



I tried to use theorems like the inscribed angle theorem, but I was not succesfull so far. Does someone know how to solve this problem?



If it is possible, I only want to use geometric arguments, and not analytical arguments. Does this result have any name?



Best wishes






geometry




geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 2 '15 at 6:28









asdasd

309114




309114












  • $begingroup$
    This theorem is true if the circles are of the same size.
    $endgroup$
    – Jack
    Jun 2 '15 at 6:51


















  • $begingroup$
    This theorem is true if the circles are of the same size.
    $endgroup$
    – Jack
    Jun 2 '15 at 6:51
















$begingroup$
This theorem is true if the circles are of the same size.
$endgroup$
– Jack
Jun 2 '15 at 6:51




$begingroup$
This theorem is true if the circles are of the same size.
$endgroup$
– Jack
Jun 2 '15 at 6:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



Let $M,N$ two points on the common tangent at opposite sides with respect to $A$, than $angle DAM = angle DEA$ because they subtend the same arc $DA$ ( $angle DAM$ is a ''limit'' angle, being the side $AM$ tangent, but, if you don't like the concept of ''limit'' angle, you can proof the same claim as a consequence of the tangent-secant theorem.).



In the same manner $angle BAN=angle BCA$.



Now note that $angle DAM$ and $angle BAN$ are opposed, and .....



I don't know if this result has a special name.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Problem Figure.



    Let the common tangent $L$ be drawn.
    Then, by the Alternate Segment Theorem:
    $$angle DEA =angle (DA,L)$$
    And similarly:
    $$angle CBA=angle (CA,L)$$
    But the latter are equal as they are vertically opposite.
    Therefore it follows that the angles
    Intercepted by $BE$ are equal and thus, the lines are parallel.




    This is a special limiting case of the Reim's Theorem, which states that lines through the points of intersection of two circles cut the circle in points that form two parallel segments.



    This is not a well-known theorem by the way.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint:



      Let $M,N$ two points on the common tangent at opposite sides with respect to $A$, than $angle DAM = angle DEA$ because they subtend the same arc $DA$ ( $angle DAM$ is a ''limit'' angle, being the side $AM$ tangent, but, if you don't like the concept of ''limit'' angle, you can proof the same claim as a consequence of the tangent-secant theorem.).



      In the same manner $angle BAN=angle BCA$.



      Now note that $angle DAM$ and $angle BAN$ are opposed, and .....



      I don't know if this result has a special name.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Hint:



        Let $M,N$ two points on the common tangent at opposite sides with respect to $A$, than $angle DAM = angle DEA$ because they subtend the same arc $DA$ ( $angle DAM$ is a ''limit'' angle, being the side $AM$ tangent, but, if you don't like the concept of ''limit'' angle, you can proof the same claim as a consequence of the tangent-secant theorem.).



        In the same manner $angle BAN=angle BCA$.



        Now note that $angle DAM$ and $angle BAN$ are opposed, and .....



        I don't know if this result has a special name.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint:



          Let $M,N$ two points on the common tangent at opposite sides with respect to $A$, than $angle DAM = angle DEA$ because they subtend the same arc $DA$ ( $angle DAM$ is a ''limit'' angle, being the side $AM$ tangent, but, if you don't like the concept of ''limit'' angle, you can proof the same claim as a consequence of the tangent-secant theorem.).



          In the same manner $angle BAN=angle BCA$.



          Now note that $angle DAM$ and $angle BAN$ are opposed, and .....



          I don't know if this result has a special name.






          share|cite|improve this answer











          $endgroup$



          Hint:



          Let $M,N$ two points on the common tangent at opposite sides with respect to $A$, than $angle DAM = angle DEA$ because they subtend the same arc $DA$ ( $angle DAM$ is a ''limit'' angle, being the side $AM$ tangent, but, if you don't like the concept of ''limit'' angle, you can proof the same claim as a consequence of the tangent-secant theorem.).



          In the same manner $angle BAN=angle BCA$.



          Now note that $angle DAM$ and $angle BAN$ are opposed, and .....



          I don't know if this result has a special name.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jun 2 '15 at 12:53

























          answered Jun 2 '15 at 9:33









          Emilio NovatiEmilio Novati

          51.6k43473




          51.6k43473























              0












              $begingroup$

              Problem Figure.



              Let the common tangent $L$ be drawn.
              Then, by the Alternate Segment Theorem:
              $$angle DEA =angle (DA,L)$$
              And similarly:
              $$angle CBA=angle (CA,L)$$
              But the latter are equal as they are vertically opposite.
              Therefore it follows that the angles
              Intercepted by $BE$ are equal and thus, the lines are parallel.




              This is a special limiting case of the Reim's Theorem, which states that lines through the points of intersection of two circles cut the circle in points that form two parallel segments.



              This is not a well-known theorem by the way.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Problem Figure.



                Let the common tangent $L$ be drawn.
                Then, by the Alternate Segment Theorem:
                $$angle DEA =angle (DA,L)$$
                And similarly:
                $$angle CBA=angle (CA,L)$$
                But the latter are equal as they are vertically opposite.
                Therefore it follows that the angles
                Intercepted by $BE$ are equal and thus, the lines are parallel.




                This is a special limiting case of the Reim's Theorem, which states that lines through the points of intersection of two circles cut the circle in points that form two parallel segments.



                This is not a well-known theorem by the way.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Problem Figure.



                  Let the common tangent $L$ be drawn.
                  Then, by the Alternate Segment Theorem:
                  $$angle DEA =angle (DA,L)$$
                  And similarly:
                  $$angle CBA=angle (CA,L)$$
                  But the latter are equal as they are vertically opposite.
                  Therefore it follows that the angles
                  Intercepted by $BE$ are equal and thus, the lines are parallel.




                  This is a special limiting case of the Reim's Theorem, which states that lines through the points of intersection of two circles cut the circle in points that form two parallel segments.



                  This is not a well-known theorem by the way.






                  share|cite|improve this answer









                  $endgroup$



                  Problem Figure.



                  Let the common tangent $L$ be drawn.
                  Then, by the Alternate Segment Theorem:
                  $$angle DEA =angle (DA,L)$$
                  And similarly:
                  $$angle CBA=angle (CA,L)$$
                  But the latter are equal as they are vertically opposite.
                  Therefore it follows that the angles
                  Intercepted by $BE$ are equal and thus, the lines are parallel.




                  This is a special limiting case of the Reim's Theorem, which states that lines through the points of intersection of two circles cut the circle in points that form two parallel segments.



                  This is not a well-known theorem by the way.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 20:02









                  ArchimedesprincipleArchimedesprinciple

                  32317




                  32317






























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