Find the area of the triangle $BDE$ if $angle ABC = 60°$, $BD = BH$, and $BE = BO = 1$.
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In the triangle $ABC$, $angle ABC = 60°$. $O$ is its circumcentre and $H$ is its orthocentre. $D$ is a point on $BC$ such that $BD = BH$. $E$ is a point on $AB$ such that $BE = BO$. If $BO = 1$, what is the area of the triangle $BDE$?
The problem is taken from here. I know the answer is $fracsqrt34$, but how can I prove it? That would mean that $BD = 1$, so the triangle $BED$ would be equilateral.
Can you give me a hint, please? Thanks!
geometry
asked Sep 9 at 11:12
Iulian Oleniuc
3929
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up vote
1
down vote
favorite
In the triangle $ABC$, $angle ABC = 60°$. $O$ is its circumcentre and $H$ is its orthocentre. $D$ is a point on $BC$ such that $BD = BH$. $E$ is a point on $AB$ such that $BE = BO$. If $BO = 1$, what is the area of the triangle $BDE$?
The problem is taken from here. I know the answer is $fracsqrt34$, but how can I prove it? That would mean that $BD = 1$, so the triangle $BED$ would be equilateral.
Can you give me a hint, please? Thanks!
geometry
asked Sep 9 at 11:12
Iulian Oleniuc
3929
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the triangle $ABC$, $angle ABC = 60°$. $O$ is its circumcentre and $H$ is its orthocentre. $D$ is a point on $BC$ such that $BD = BH$. $E$ is a point on $AB$ such that $BE = BO$. If $BO = 1$, what is the area of the triangle $BDE$?
The problem is taken from here. I know the answer is $fracsqrt34$, but how can I prove it? That would mean that $BD = 1$, so the triangle $BED$ would be equilateral.
Can you give me a hint, please? Thanks!
geometry
asked Sep 9 at 11:12
Iulian Oleniuc
3929
In the triangle $ABC$, $angle ABC = 60°$. $O$ is its circumcentre and $H$ is its orthocentre. $D$ is a point on $BC$ such that $BD = BH$. $E$ is a point on $AB$ such that $BE = BO$. If $BO = 1$, what is the area of the triangle $BDE$?
The problem is taken from here. I know the answer is $fracsqrt34$, but how can I prove it? That would mean that $BD = 1$, so the triangle $BED$ would be equilateral.
Can you give me a hint, please? Thanks!
geometry
geometry
asked Sep 9 at 11:12
Iulian Oleniuc
3929
asked Sep 9 at 11:12
Iulian Oleniuc
3929
asked Sep 9 at 11:12
Iulian Oleniuc
3929
asked Sep 9 at 11:12
Iulian Oleniuc
3929
asked Sep 9 at 11:12
Iulian Oleniuc
3929
3929
add a comment |Â
add a comment |Â
1 Answer
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Let $R'$ be the radius of the circumcircle of $triangleBCH$.
By the law of sines,
$$1=fracBC2sinangleBAC,quad R'=fracBC2sinangleBHCtag1$$
Since $angleBHC=180^circ-angleBAC$, we have
$$sinangleBHC=sinangleBACtag2$$
From $(1),(2)$, we have
$$R'=1$$
So, $$BD=BH=2R'sinangleHCB=2sin(90^circ-angleABC)=1$$
from which
$$[BED]=frac 12times BDtimes BEtimessin60^circ=frac 12times 1times 1timesfracsqrt 32=fracsqrt 34$$
follows.
answered Sep 9 at 12:27
mathlove
87.8k877209
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $R'$ be the radius of the circumcircle of $triangleBCH$.
By the law of sines,
$$1=fracBC2sinangleBAC,quad R'=fracBC2sinangleBHCtag1$$
Since $angleBHC=180^circ-angleBAC$, we have
$$sinangleBHC=sinangleBACtag2$$
From $(1),(2)$, we have
$$R'=1$$
So, $$BD=BH=2R'sinangleHCB=2sin(90^circ-angleABC)=1$$
from which
$$[BED]=frac 12times BDtimes BEtimessin60^circ=frac 12times 1times 1timesfracsqrt 32=fracsqrt 34$$
follows.
answered Sep 9 at 12:27
mathlove
87.8k877209
add a comment |Â
up vote
2
down vote
accepted
Let $R'$ be the radius of the circumcircle of $triangleBCH$.
By the law of sines,
$$1=fracBC2sinangleBAC,quad R'=fracBC2sinangleBHCtag1$$
Since $angleBHC=180^circ-angleBAC$, we have
$$sinangleBHC=sinangleBACtag2$$
From $(1),(2)$, we have
$$R'=1$$
So, $$BD=BH=2R'sinangleHCB=2sin(90^circ-angleABC)=1$$
from which
$$[BED]=frac 12times BDtimes BEtimessin60^circ=frac 12times 1times 1timesfracsqrt 32=fracsqrt 34$$
follows.
answered Sep 9 at 12:27
mathlove
87.8k877209
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $R'$ be the radius of the circumcircle of $triangleBCH$.
By the law of sines,
$$1=fracBC2sinangleBAC,quad R'=fracBC2sinangleBHCtag1$$
Since $angleBHC=180^circ-angleBAC$, we have
$$sinangleBHC=sinangleBACtag2$$
From $(1),(2)$, we have
$$R'=1$$
So, $$BD=BH=2R'sinangleHCB=2sin(90^circ-angleABC)=1$$
from which
$$[BED]=frac 12times BDtimes BEtimessin60^circ=frac 12times 1times 1timesfracsqrt 32=fracsqrt 34$$
follows.
answered Sep 9 at 12:27
mathlove
87.8k877209
Let $R'$ be the radius of the circumcircle of $triangleBCH$.
By the law of sines,
$$1=fracBC2sinangleBAC,quad R'=fracBC2sinangleBHCtag1$$
Since $angleBHC=180^circ-angleBAC$, we have
$$sinangleBHC=sinangleBACtag2$$
From $(1),(2)$, we have
$$R'=1$$
So, $$BD=BH=2R'sinangleHCB=2sin(90^circ-angleABC)=1$$
from which
$$[BED]=frac 12times BDtimes BEtimessin60^circ=frac 12times 1times 1timesfracsqrt 32=fracsqrt 34$$
follows.
answered Sep 9 at 12:27
mathlove
87.8k877209
answered Sep 9 at 12:27
mathlove
87.8k877209
answered Sep 9 at 12:27
mathlove
87.8k877209
answered Sep 9 at 12:27
mathlove
87.8k877209
87.8k877209
add a comment |Â
add a comment |Â
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