Prove that $BD$ bisects $angle ABC$












7












$begingroup$



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






geometry euclidean-geometry triangles







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Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    Apr 20 at 4:06










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    Apr 20 at 4:20










  • $begingroup$
    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
    Apr 20 at 4:24
















7












$begingroup$



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






geometry euclidean-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.







$endgroup$












  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    Apr 20 at 4:06










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    Apr 20 at 4:20










  • $begingroup$
    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
    Apr 20 at 4:24














7












7








7





$begingroup$



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






geometry euclidean-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.







$endgroup$





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 euclidean-geometry triangles




geometry euclidean-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 Apr 20 at 12:12









user21820

40.4k544163




40.4k544163






New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Apr 20 at 4:01









Pushpa KumariPushpa Kumari

365




365




New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    Apr 20 at 4:06










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    Apr 20 at 4:20










  • $begingroup$
    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
    Apr 20 at 4:24


















  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    Apr 20 at 4:06










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    Apr 20 at 4:20










  • $begingroup$
    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
    Apr 20 at 4:24
















$begingroup$
Start by drawing a diagram and showing all the given information.
$endgroup$
– 1123581321
Apr 20 at 4:06




$begingroup$
Start by drawing a diagram and showing all the given information.
$endgroup$
– 1123581321
Apr 20 at 4:06












$begingroup$
@PushpaKumari just provide a link to your image, someone will be willing to edit it.
$endgroup$
– Quang Hoang
Apr 20 at 4:20




$begingroup$
@PushpaKumari just provide a link to your image, someone will be willing to edit it.
$endgroup$
– Quang Hoang
Apr 20 at 4:20












$begingroup$
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
Apr 20 at 4:24




$begingroup$
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
Apr 20 at 4:24










3 Answers
3






active

oldest

votes


















5












$begingroup$

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
    Apr 20 at 6:20










  • $begingroup$
    Mind blown 😱😱
    $endgroup$
    – Pushpa Kumari
    Apr 20 at 8:10



















4












$begingroup$

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









$endgroup$





















    3












    $begingroup$

    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









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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
        Apr 20 at 6:20










      • $begingroup$
        Mind blown 😱😱
        $endgroup$
        – Pushpa Kumari
        Apr 20 at 8:10
















      5












      $begingroup$

      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
        Apr 20 at 6:20










      • $begingroup$
        Mind blown 😱😱
        $endgroup$
        – Pushpa Kumari
        Apr 20 at 8:10














      5












      5








      5





      $begingroup$

      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 Apr 20 at 6:04









      siroussirous

      1,8131514




      1,8131514












      • $begingroup$
        +1. Thinking out of box.
        $endgroup$
        – farruhota
        Apr 20 at 6:20










      • $begingroup$
        Mind blown 😱😱
        $endgroup$
        – Pushpa Kumari
        Apr 20 at 8:10


















      • $begingroup$
        +1. Thinking out of box.
        $endgroup$
        – farruhota
        Apr 20 at 6:20










      • $begingroup$
        Mind blown 😱😱
        $endgroup$
        – Pushpa Kumari
        Apr 20 at 8:10
















      $begingroup$
      +1. Thinking out of box.
      $endgroup$
      – farruhota
      Apr 20 at 6:20




      $begingroup$
      +1. Thinking out of box.
      $endgroup$
      – farruhota
      Apr 20 at 6:20












      $begingroup$
      Mind blown 😱😱
      $endgroup$
      – Pushpa Kumari
      Apr 20 at 8:10




      $begingroup$
      Mind blown 😱😱
      $endgroup$
      – Pushpa Kumari
      Apr 20 at 8:10











      4












      $begingroup$

      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









      $endgroup$


















        4












        $begingroup$

        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









        $endgroup$
















          4












          4








          4





          $begingroup$

          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 Apr 20 at 4:41









          farruhotafarruhota

          22.5k2942




          22.5k2942























              3












              $begingroup$

              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









              $endgroup$


















                3












                $begingroup$

                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









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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 Apr 20 at 4:59









                  Quang HoangQuang Hoang

                  13.4k1233




                  13.4k1233






















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










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                      Cáceres

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                      Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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