Let $AD$ be the angle bisector of angle $A$ in $Delta ABC$. Prove that $BD = BC$ . $fracABAB + AC$How to show angle bisector in triangle ABC?Finding the Perimiter of a right Triangle given an interior angle bisector and exterior angle bisectorIn this figure, prove that $H$ lies on the circle iff it lies on the perpendicular bisector of $Delta ABC$Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrentFinding an angle in a triangle, given the angle bisector and some conditions.How to prove that these lines are concurrent?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 $Delta ABC, $ $K$ and $L$ are points on $BC$. $AL$ is the bisector of $angle KAC$. $KLtimes BC=BKtimes CL$. Find $angle BAL$.Intersection of angle bisector and perpendicular bisector of opposite sideAngle bisector contains the Nine Point Centre
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Let $AD$ be the angle bisector of angle $A$ in $Delta ABC$. Prove that $BD = BC$ . $fracABAB + AC$
How to show angle bisector in triangle ABC?Finding the Perimiter of a right Triangle given an interior angle bisector and exterior angle bisectorIn this figure, prove that $H$ lies on the circle iff it lies on the perpendicular bisector of $Delta ABC$Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrentFinding an angle in a triangle, given the angle bisector and some conditions.How to prove that these lines are concurrent?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 $Delta ABC, $ $K$ and $L$ are points on $BC$. $AL$ is the bisector of $angle KAC$. $KLtimes BC=BKtimes CL$. Find $angle BAL$.Intersection of angle bisector and perpendicular bisector of opposite sideAngle bisector contains the Nine Point Centre
$begingroup$
Let $AD$ be the angle bisector of angle $A$ in $Delta ABC$. Prove that $$BD = BC cdot fracABAB + AC$$
Hello,
I was doing some geometry and got stuck in this question. I tried using the angle bisector theorem and I know it will be used somewhere in this problem but can’t really get it right. Can you please help me with this question? I would be grateful if you did.
Thanks.
geometry euclidean-geometry triangles
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
asked Mar 28 at 15:36
Vasu090Vasu090
244
$endgroup$
add a comment |
$begingroup$
Let $AD$ be the angle bisector of angle $A$ in $Delta ABC$. Prove that $$BD = BC cdot fracABAB + AC$$
Hello,
I was doing some geometry and got stuck in this question. I tried using the angle bisector theorem and I know it will be used somewhere in this problem but can’t really get it right. Can you please help me with this question? I would be grateful if you did.
Thanks.
geometry euclidean-geometry triangles
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
asked Mar 28 at 15:36
Vasu090Vasu090
244
$endgroup$
$begingroup$
Welcome to Math.SE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Brian
Mar 28 at 15:38
add a comment |
$begingroup$
Let $AD$ be the angle bisector of angle $A$ in $Delta ABC$. Prove that $$BD = BC cdot fracABAB + AC$$
Hello,
I was doing some geometry and got stuck in this question. I tried using the angle bisector theorem and I know it will be used somewhere in this problem but can’t really get it right. Can you please help me with this question? I would be grateful if you did.
Thanks.
geometry euclidean-geometry triangles
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
asked Mar 28 at 15:36
Vasu090Vasu090
244
$endgroup$
Let $AD$ be the angle bisector of angle $A$ in $Delta ABC$. Prove that $$BD = BC cdot fracABAB + AC$$
Hello,
I was doing some geometry and got stuck in this question. I tried using the angle bisector theorem and I know it will be used somewhere in this problem but can’t really get it right. Can you please help me with this question? I would be grateful if you did.
Thanks.
geometry euclidean-geometry triangles
geometry euclidean-geometry triangles
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
asked Mar 28 at 15:36
Vasu090Vasu090
244
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
asked Mar 28 at 15:36
Vasu090Vasu090
244
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
edited Mar 28 at 16:08
Maria Mazur
49.5k1361124
49.5k1361124
asked Mar 28 at 15:36
Vasu090Vasu090
244
asked Mar 28 at 15:36
Vasu090Vasu090
244
asked Mar 28 at 15:36
Vasu090Vasu090
244
244
$begingroup$
Welcome to Math.SE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Brian
Mar 28 at 15:38
add a comment |
$begingroup$
Welcome to Math.SE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Brian
Mar 28 at 15:38
$begingroup$
Welcome to Math.SE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Brian
Mar 28 at 15:38
$begingroup$
Welcome to Math.SE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Brian
Mar 28 at 15:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the theorem about the internal bisector, we get
$$fracBDDC=fraccb$$ so we get
$$BD=DCcdot fraccb$$
Using that $$DC=a-BD$$
We get
$$BD=(a-BD)cdot fraccb$$
and we obtain
$$BD(1+fraccb)=fracacb$$
Can you finish?
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
Extend side $AB$ across $A$ for $AC$, we get new point $E$. Then $angle AEC = angle BAD$ so $EC||AD$. By Thales theorem we obtain: $$ BDover BA= BCover BEimplies BD = BAcdot BCover BE =ABcdot BCover AB+AC $$
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Using the theorem about the internal bisector, we get
$$fracBDDC=fraccb$$ so we get
$$BD=DCcdot fraccb$$
Using that $$DC=a-BD$$
We get
$$BD=(a-BD)cdot fraccb$$
and we obtain
$$BD(1+fraccb)=fracacb$$
Can you finish?
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
Using the theorem about the internal bisector, we get
$$fracBDDC=fraccb$$ so we get
$$BD=DCcdot fraccb$$
Using that $$DC=a-BD$$
We get
$$BD=(a-BD)cdot fraccb$$
and we obtain
$$BD(1+fraccb)=fracacb$$
Can you finish?
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
Using the theorem about the internal bisector, we get
$$fracBDDC=fraccb$$ so we get
$$BD=DCcdot fraccb$$
Using that $$DC=a-BD$$
We get
$$BD=(a-BD)cdot fraccb$$
and we obtain
$$BD(1+fraccb)=fracacb$$
Can you finish?
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
Using the theorem about the internal bisector, we get
$$fracBDDC=fraccb$$ so we get
$$BD=DCcdot fraccb$$
Using that $$DC=a-BD$$
We get
$$BD=(a-BD)cdot fraccb$$
and we obtain
$$BD(1+fraccb)=fracacb$$
Can you finish?
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 28 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
add a comment |
add a comment |
$begingroup$
Extend side $AB$ across $A$ for $AC$, we get new point $E$. Then $angle AEC = angle BAD$ so $EC||AD$. By Thales theorem we obtain: $$ BDover BA= BCover BEimplies BD = BAcdot BCover BE =ABcdot BCover AB+AC $$
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
add a comment |
$begingroup$
Extend side $AB$ across $A$ for $AC$, we get new point $E$. Then $angle AEC = angle BAD$ so $EC||AD$. By Thales theorem we obtain: $$ BDover BA= BCover BEimplies BD = BAcdot BCover BE =ABcdot BCover AB+AC $$
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
add a comment |
$begingroup$
Extend side $AB$ across $A$ for $AC$, we get new point $E$. Then $angle AEC = angle BAD$ so $EC||AD$. By Thales theorem we obtain: $$ BDover BA= BCover BEimplies BD = BAcdot BCover BE =ABcdot BCover AB+AC $$
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
Extend side $AB$ across $A$ for $AC$, we get new point $E$. Then $angle AEC = angle BAD$ so $EC||AD$. By Thales theorem we obtain: $$ BDover BA= BCover BEimplies BD = BAcdot BCover BE =ABcdot BCover AB+AC $$
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
edited Mar 28 at 17:46
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
answered Mar 28 at 16:07
Maria MazurMaria Mazur
49.5k1361124
49.5k1361124
add a comment |
add a comment |
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$begingroup$
Welcome to Math.SE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Brian
Mar 28 at 15:38