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







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












$begingroup$


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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1308858%2ftwo-tangent-circles-and-parallel-lines%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1308858%2ftwo-tangent-circles-and-parallel-lines%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      An IMO inspired problem

                      Image
                      14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                      Read more

                      Management

                      Image
                      Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                      Read more

                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

                      Image
                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
                      Read more