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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.

geometry triangles

share|cite|improve this question

edited 1 hour ago

Pushpa Kumari

asked 2 hours ago

Pushpa KumariPushpa Kumari

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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.

geometry triangles

share|cite|improve this question

edited 1 hour ago

Pushpa Kumari

asked 2 hours ago

Pushpa KumariPushpa Kumari

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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.

geometry triangles

share|cite|improve this question

edited 1 hour ago

Pushpa Kumari

asked 2 hours ago

Pushpa KumariPushpa Kumari

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.

$endgroup$

share|cite|improve this question

edited 1 hour ago

Pushpa Kumari

asked 2 hours ago

Pushpa KumariPushpa Kumari

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.

share|cite|improve this question

edited 1 hour ago

Pushpa Kumari

asked 2 hours ago

Pushpa KumariPushpa Kumari

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.

asked 2 hours ago

Pushpa KumariPushpa Kumari

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.

asked 2 hours ago

Pushpa KumariPushpa Kumari

334

3 Answers
3

active

oldest

votes

1

$begingroup$

Refer to the figure:

$hspace{2cm}$

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

answered 1 hour ago

farruhotafarruhota

22.3k2942

$endgroup$

add a comment | 

1

$begingroup$

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

answered 1 hour ago

Quang HoangQuang Hoang

13.3k1233

$endgroup$

add a comment | 

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 $AE=frac {1}{2}DB=frac {1}{2}AF$. 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

answered 6 mins ago

siroussirous

1,7481514

$endgroup$

add a comment | 

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1

$begingroup$

Refer to the figure:

$hspace{2cm}$

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

answered 1 hour ago

farruhotafarruhota

22.3k2942

$endgroup$

add a comment | 

1

$begingroup$

Refer to the figure:

$hspace{2cm}$

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

answered 1 hour ago

farruhotafarruhota

22.3k2942

$endgroup$

add a comment | 

$begingroup$

Refer to the figure:

$hspace{2cm}$

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

answered 1 hour ago

farruhotafarruhota

22.3k2942

$endgroup$

share|cite|improve this answer

answered 1 hour ago

farruhotafarruhota

22.3k2942

answered 1 hour ago

farruhotafarruhota

22.3k2942

answered 1 hour ago

farruhotafarruhota

22.3k2942

1

$begingroup$

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

answered 1 hour ago

Quang HoangQuang Hoang

13.3k1233

$endgroup$

add a comment | 

1

$begingroup$

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

answered 1 hour ago

Quang HoangQuang Hoang

13.3k1233

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

Quang HoangQuang Hoang

13.3k1233

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Quang HoangQuang Hoang

13.3k1233

answered 1 hour ago

Quang HoangQuang Hoang

13.3k1233

answered 1 hour ago

Quang HoangQuang Hoang

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 $AE=frac {1}{2}DB=frac {1}{2}AF$. 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

answered 6 mins ago

siroussirous

1,7481514

$endgroup$

add a comment | 

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 $AE=frac {1}{2}DB=frac {1}{2}AF$. 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

answered 6 mins ago

siroussirous

1,7481514

$endgroup$

add a comment | 

$begingroup$

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

answered 6 mins ago

siroussirous

1,7481514

$endgroup$

share|cite

answered 6 mins ago

siroussirous

1,7481514

answered 6 mins ago

siroussirous

1,7481514

answered 6 mins ago

siroussirous

1,7481514