Prove that BD bisects angle ABC












6












$begingroup$



Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac{1}{2}BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here










share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    2 hours ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    2 hours ago
















6












$begingroup$



Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac{1}{2}BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here










share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    2 hours ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    2 hours ago














6












6








6





$begingroup$



Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac{1}{2}BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here










share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac{1}{2}BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here







geometry triangles






share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







Pushpa Kumari













New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









Pushpa KumariPushpa Kumari

334




334




New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    2 hours ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    2 hours ago


















  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    2 hours ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    2 hours ago
















$begingroup$
Start by drawing a diagram and showing all the given information.
$endgroup$
– 1123581321
2 hours ago




$begingroup$
Start by drawing a diagram and showing all the given information.
$endgroup$
– 1123581321
2 hours ago












$begingroup$
@PushpaKumari just provide a link to your image, someone will be willing to edit it.
$endgroup$
– Quang Hoang
2 hours ago




$begingroup$
@PushpaKumari just provide a link to your image, someone will be willing to edit it.
$endgroup$
– Quang Hoang
2 hours ago












$begingroup$
To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
$endgroup$
– 1123581321
2 hours ago




$begingroup$
To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
$endgroup$
– 1123581321
2 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Refer to the figure:



$hspace{2cm}$enter image description here



From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
$$frac{x}{y}=frac{y-z}{2x} Rightarrow 2x^2=y^2-zy (1)$$
From the right $Delta BCD$:
$$z^2+y^2=(2x)^2 (2)$$
Now substitute $(1)$ to $(2)$:
$$z^2+y^2=2(y^2-zy) Rightarrow \
(y-z)^2=2z^2 Rightarrow \
y-z=zsqrt{2} Rightarrow \
frac{y-z}{z}=frac{ysqrt{2}}{y},$$

which is consistent with the angle bisector theorem.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    enter image description here



    Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



    First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



    Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
    Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      A simple geometric solution:



      Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac {1}{2}DB=frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        +1. Thinking out of box.
        $endgroup$
        – farruhota
        28 mins ago












      Your Answer








      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
      });


      }
      });






      Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3194309%2fprove-that-bd-bisects-angle-abc%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Refer to the figure:



      $hspace{2cm}$enter image description here



      From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
      $$frac{x}{y}=frac{y-z}{2x} Rightarrow 2x^2=y^2-zy (1)$$
      From the right $Delta BCD$:
      $$z^2+y^2=(2x)^2 (2)$$
      Now substitute $(1)$ to $(2)$:
      $$z^2+y^2=2(y^2-zy) Rightarrow \
      (y-z)^2=2z^2 Rightarrow \
      y-z=zsqrt{2} Rightarrow \
      frac{y-z}{z}=frac{ysqrt{2}}{y},$$

      which is consistent with the angle bisector theorem.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Refer to the figure:



        $hspace{2cm}$enter image description here



        From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
        $$frac{x}{y}=frac{y-z}{2x} Rightarrow 2x^2=y^2-zy (1)$$
        From the right $Delta BCD$:
        $$z^2+y^2=(2x)^2 (2)$$
        Now substitute $(1)$ to $(2)$:
        $$z^2+y^2=2(y^2-zy) Rightarrow \
        (y-z)^2=2z^2 Rightarrow \
        y-z=zsqrt{2} Rightarrow \
        frac{y-z}{z}=frac{ysqrt{2}}{y},$$

        which is consistent with the angle bisector theorem.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Refer to the figure:



          $hspace{2cm}$enter image description here



          From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
          $$frac{x}{y}=frac{y-z}{2x} Rightarrow 2x^2=y^2-zy (1)$$
          From the right $Delta BCD$:
          $$z^2+y^2=(2x)^2 (2)$$
          Now substitute $(1)$ to $(2)$:
          $$z^2+y^2=2(y^2-zy) Rightarrow \
          (y-z)^2=2z^2 Rightarrow \
          y-z=zsqrt{2} Rightarrow \
          frac{y-z}{z}=frac{ysqrt{2}}{y},$$

          which is consistent with the angle bisector theorem.






          share|cite|improve this answer









          $endgroup$



          Refer to the figure:



          $hspace{2cm}$enter image description here



          From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
          $$frac{x}{y}=frac{y-z}{2x} Rightarrow 2x^2=y^2-zy (1)$$
          From the right $Delta BCD$:
          $$z^2+y^2=(2x)^2 (2)$$
          Now substitute $(1)$ to $(2)$:
          $$z^2+y^2=2(y^2-zy) Rightarrow \
          (y-z)^2=2z^2 Rightarrow \
          y-z=zsqrt{2} Rightarrow \
          frac{y-z}{z}=frac{ysqrt{2}}{y},$$

          which is consistent with the angle bisector theorem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          farruhotafarruhota

          22.3k2942




          22.3k2942























              2












              $begingroup$

              enter image description here



              Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



              First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



              Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
              Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                enter image description here



                Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  enter image description here



                  Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                  First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                  Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                  Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






                  share|cite|improve this answer









                  $endgroup$



                  enter image description here



                  Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                  First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                  Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                  Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Quang HoangQuang Hoang

                  13.3k1233




                  13.3k1233























                      1












                      $begingroup$

                      A simple geometric solution:



                      Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac {1}{2}DB=frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        +1. Thinking out of box.
                        $endgroup$
                        – farruhota
                        28 mins ago
















                      1












                      $begingroup$

                      A simple geometric solution:



                      Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac {1}{2}DB=frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        +1. Thinking out of box.
                        $endgroup$
                        – farruhota
                        28 mins ago














                      1












                      1








                      1





                      $begingroup$

                      A simple geometric solution:



                      Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac {1}{2}DB=frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






                      share|cite|improve this answer









                      $endgroup$



                      A simple geometric solution:



                      Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac {1}{2}DB=frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 44 mins ago









                      siroussirous

                      1,7581514




                      1,7581514












                      • $begingroup$
                        +1. Thinking out of box.
                        $endgroup$
                        – farruhota
                        28 mins ago


















                      • $begingroup$
                        +1. Thinking out of box.
                        $endgroup$
                        – farruhota
                        28 mins ago
















                      $begingroup$
                      +1. Thinking out of box.
                      $endgroup$
                      – farruhota
                      28 mins ago




                      $begingroup$
                      +1. Thinking out of box.
                      $endgroup$
                      – farruhota
                      28 mins ago










                      Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.













                      Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.












                      Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.
















                      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%2f3194309%2fprove-that-bd-bisects-angle-abc%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

                      Liste der Baudenkmale in Friedland (Mecklenburg)

                      Single-Malt-Whisky

                      Czorneboh