Prove that BD bisects angle ABC Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A geometry problem that involves congruence of triangles.Prove that: $S_XYZgeq frac14S_ABC$Prove that angle ACB > angle ABD.Let $D, E, F$ be the feet of the altitudes from $A, B, C$ in $triangle ABC$. Prove that the perpendicular bisector of $EF$ also bisects $BC$.In the following figure, prove that $AC$ bisects $GH$.In triangle $ABC$ find angle $angle BAC$ given that…Show that the altitude bisects the corresponding angleAngle bisector contains the Nine Point CentreProve sum of angles in problem involving bisectors in a given triangleHow to solve for $angle BDC$ given the information of other angles in the picture

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Prove that BD bisects angle ABC

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Prove that BD bisects angle ABC



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A geometry problem that involves congruence of triangles.Prove that: $S_XYZgeq frac14S_ABC$Prove that angle ACB > angle ABD.Let $D, E, F$ be the feet of the altitudes from $A, B, C$ in $triangle ABC$. Prove that the perpendicular bisector of $EF$ also bisects $BC$.In the following figure, prove that $AC$ bisects $GH$.In triangle $ABC$ find angle $angle BAC$ given that…Show that the altitude bisects the corresponding angleAngle bisector contains the Nine Point CentreProve sum of angles in problem involving bisectors in a given triangleHow to solve for $angle BDC$ given the information of other angles in the picture










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 $AE=frac12BD$, 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










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  • $begingroup$
    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
    1 hour 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
    1 hour 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 $AE=frac12BD$, 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
    1 hour 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
    1 hour 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 $AE=frac12BD$, 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 $AE=frac12BD$, 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 1 hour ago







Pushpa Kumari













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asked 2 hours ago









Pushpa KumariPushpa Kumari

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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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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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
1 hour ago




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










3 Answers
3






active

oldest

votes


















1












$begingroup$

Refer to the figure:



$hspace2cm$enter image description here



From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
$$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
From the right $Delta BCD$:
$$z^2+y^2=(2x)^2 (2)$$
Now substitute $(1)$ to $(2)$:
$$z^2+y^2=2(y^2-zy) Rightarrow \
(y-z)^2=2z^2 Rightarrow \
y-z=zsqrt2 Rightarrow \
fracy-zz=fracysqrt2y,$$

which is consistent with the angle bisector theorem.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    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









    $endgroup$




















      0












      $begingroup$

      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 $AE=frac 12DB=frac 12AF$. 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









      $endgroup$













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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Refer to the figure:



        $hspace2cm$enter image description here



        From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
        $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
        From the right $Delta BCD$:
        $$z^2+y^2=(2x)^2 (2)$$
        Now substitute $(1)$ to $(2)$:
        $$z^2+y^2=2(y^2-zy) Rightarrow \
        (y-z)^2=2z^2 Rightarrow \
        y-z=zsqrt2 Rightarrow \
        fracy-zz=fracysqrt2y,$$

        which is consistent with the angle bisector theorem.






        share|cite|improve this answer









        $endgroup$

















          1












          $begingroup$

          Refer to the figure:



          $hspace2cm$enter image description here



          From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
          $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
          From the right $Delta BCD$:
          $$z^2+y^2=(2x)^2 (2)$$
          Now substitute $(1)$ to $(2)$:
          $$z^2+y^2=2(y^2-zy) Rightarrow \
          (y-z)^2=2z^2 Rightarrow \
          y-z=zsqrt2 Rightarrow \
          fracy-zz=fracysqrt2y,$$

          which is consistent with the angle bisector theorem.






          share|cite|improve this answer









          $endgroup$















            1












            1








            1





            $begingroup$

            Refer to the figure:



            $hspace2cm$enter image description here



            From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
            $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
            From the right $Delta BCD$:
            $$z^2+y^2=(2x)^2 (2)$$
            Now substitute $(1)$ to $(2)$:
            $$z^2+y^2=2(y^2-zy) Rightarrow \
            (y-z)^2=2z^2 Rightarrow \
            y-z=zsqrt2 Rightarrow \
            fracy-zz=fracysqrt2y,$$

            which is consistent with the angle bisector theorem.






            share|cite|improve this answer









            $endgroup$



            Refer to the figure:



            $hspace2cm$enter image description here



            From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
            $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
            From the right $Delta BCD$:
            $$z^2+y^2=(2x)^2 (2)$$
            Now substitute $(1)$ to $(2)$:
            $$z^2+y^2=2(y^2-zy) Rightarrow \
            (y-z)^2=2z^2 Rightarrow \
            y-z=zsqrt2 Rightarrow \
            fracy-zz=fracysqrt2y,$$

            which is consistent with the angle bisector theorem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            farruhotafarruhota

            22.3k2942




            22.3k2942





















                1












                $begingroup$

                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












                  $begingroup$

                  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












                    1








                    1





                    $begingroup$

                    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





















                        0












                        $begingroup$

                        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 $AE=frac 12DB=frac 12AF$. 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









                        $endgroup$

















                          0












                          $begingroup$

                          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 $AE=frac 12DB=frac 12AF$. 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









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            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 $AE=frac 12DB=frac 12AF$. 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









                            $endgroup$



                            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 $AE=frac 12DB=frac 12AF$. 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












                            share|cite



                            share|cite










                            answered 5 mins ago









                            siroussirous

                            1,7481514




                            1,7481514




















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









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                                Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

                                2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

                                Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e