Prove that BD bisects angle ABC
$begingroup$
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
asked 2 hours ago
Pushpa KumariPushpa Kumari
334
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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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$endgroup$
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
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
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
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
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.
$endgroup$
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
geometry triangles
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.
edited 2 hours 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
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
farruhotafarruhota
22.3k2942
$endgroup$
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 $angle ABC$.
answered 1 hour ago
Quang HoangQuang Hoang
13.3k1233
$endgroup$
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$.
answered 39 mins ago
siroussirous
1,7581514
$endgroup$
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 2 hours 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.
answered 2 hours 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.
answered 2 hours ago
farruhotafarruhota
22.3k2942
$endgroup$
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.
answered 2 hours ago
farruhotafarruhota
22.3k2942
answered 2 hours ago
farruhotafarruhota
22.3k2942
answered 2 hours ago
farruhotafarruhota
22.3k2942
answered 2 hours ago
farruhotafarruhota
22.3k2942
22.3k2942
add a comment |
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 $angle ABC$.
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 $angle ABC$.
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 $angle ABC$.
answered 1 hour ago
Quang HoangQuang Hoang
13.3k1233
$endgroup$
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 $angle ABC$.
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
answered 1 hour ago
Quang HoangQuang Hoang
13.3k1233
13.3k1233
add a comment |
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$.
answered 39 mins ago
siroussirous
1,7581514
$endgroup$
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
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$.
answered 39 mins ago
siroussirous
1,7581514
$endgroup$
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
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$.
answered 39 mins ago
siroussirous
1,7581514
$endgroup$
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$.
answered 39 mins ago
siroussirous
1,7581514
answered 39 mins ago
siroussirous
1,7581514
answered 39 mins ago
siroussirous
1,7581514
answered 39 mins ago
siroussirous
1,7581514
1,7581514
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
add a comment |
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
$begingroup$
+1. Thinking out of box.
$endgroup$
– farruhota
23 mins ago
add a comment |
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.
Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.
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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
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