Prove that in $triangle ABC$ ,$QPparallel BC$
up vote
1
down vote
favorite
In the triangle $triangle ABC$ , the point $M$ is between $B$ and $C$. And also the lines $MP$ and $MQ$ are bisectors of $angle AMC$ and $angle AMB$. It means that: $$angle AMP=angle PMC$$
$$angle AMQ=angle QMB$$
and
$$BM=MC$$
So now the puzzle tells us to prove that:$$QPparallel BC$$
So I know Thales's theorem and all relations between the similiar triangles. But I can't find any pairs of similiar triangles or any parallel lines to use the Thales's theorem!
Please help me proving $QPparallel BC$.
geometry euclidean-geometry triangle
edited Nov 16 at 10:13
Batominovski
31.6k23188
asked Nov 16 at 10:09
user602338
1326
add a comment |
up vote
1
down vote
favorite
In the triangle $triangle ABC$ , the point $M$ is between $B$ and $C$. And also the lines $MP$ and $MQ$ are bisectors of $angle AMC$ and $angle AMB$. It means that: $$angle AMP=angle PMC$$
$$angle AMQ=angle QMB$$
and
$$BM=MC$$
So now the puzzle tells us to prove that:$$QPparallel BC$$
So I know Thales's theorem and all relations between the similiar triangles. But I can't find any pairs of similiar triangles or any parallel lines to use the Thales's theorem!
Please help me proving $QPparallel BC$.
geometry euclidean-geometry triangle
edited Nov 16 at 10:13
Batominovski
31.6k23188
asked Nov 16 at 10:09
user602338
1326
Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect.
– Eevee Trainer
Nov 16 at 10:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the triangle $triangle ABC$ , the point $M$ is between $B$ and $C$. And also the lines $MP$ and $MQ$ are bisectors of $angle AMC$ and $angle AMB$. It means that: $$angle AMP=angle PMC$$
$$angle AMQ=angle QMB$$
and
$$BM=MC$$
So now the puzzle tells us to prove that:$$QPparallel BC$$
So I know Thales's theorem and all relations between the similiar triangles. But I can't find any pairs of similiar triangles or any parallel lines to use the Thales's theorem!
Please help me proving $QPparallel BC$.
geometry euclidean-geometry triangle
edited Nov 16 at 10:13
Batominovski
31.6k23188
asked Nov 16 at 10:09
user602338
1326
In the triangle $triangle ABC$ , the point $M$ is between $B$ and $C$. And also the lines $MP$ and $MQ$ are bisectors of $angle AMC$ and $angle AMB$. It means that: $$angle AMP=angle PMC$$
$$angle AMQ=angle QMB$$
and
$$BM=MC$$
So now the puzzle tells us to prove that:$$QPparallel BC$$
So I know Thales's theorem and all relations between the similiar triangles. But I can't find any pairs of similiar triangles or any parallel lines to use the Thales's theorem!
Please help me proving $QPparallel BC$.
geometry euclidean-geometry triangle
geometry euclidean-geometry triangle
edited Nov 16 at 10:13
Batominovski
31.6k23188
asked Nov 16 at 10:09
user602338
1326
edited Nov 16 at 10:13
Batominovski
31.6k23188
asked Nov 16 at 10:09
user602338
1326
edited Nov 16 at 10:13
Batominovski
31.6k23188
edited Nov 16 at 10:13
Batominovski
31.6k23188
edited Nov 16 at 10:13
Batominovski
31.6k23188
31.6k23188
asked Nov 16 at 10:09
user602338
1326
asked Nov 16 at 10:09
user602338
1326
asked Nov 16 at 10:09
user602338
1326
1326
Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect.
– Eevee Trainer
Nov 16 at 10:13
add a comment |
Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect.
– Eevee Trainer
Nov 16 at 10:13
Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect.
– Eevee Trainer
Nov 16 at 10:13
Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect.
– Eevee Trainer
Nov 16 at 10:13
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
By the Angle Bisector Theorem, $$frac{AQ}{QB}=frac{AM}{MB}text{ and }frac{AP}{PC}=frac{AM}{MC},.$$
Since $M$ is the midpoint of $BC$, we have $MB=MC$, whence
$$frac{AQ}{QB}=frac{AP}{PC},.$$
Therefore, $PQparallel BC$.
answered Nov 16 at 10:15
Batominovski
31.6k23188
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
|
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
By the Angle Bisector Theorem, $$frac{AQ}{QB}=frac{AM}{MB}text{ and }frac{AP}{PC}=frac{AM}{MC},.$$
Since $M$ is the midpoint of $BC$, we have $MB=MC$, whence
$$frac{AQ}{QB}=frac{AP}{PC},.$$
Therefore, $PQparallel BC$.
answered Nov 16 at 10:15
Batominovski
31.6k23188
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
|
show 5 more comments
up vote
4
down vote
By the Angle Bisector Theorem, $$frac{AQ}{QB}=frac{AM}{MB}text{ and }frac{AP}{PC}=frac{AM}{MC},.$$
Since $M$ is the midpoint of $BC$, we have $MB=MC$, whence
$$frac{AQ}{QB}=frac{AP}{PC},.$$
Therefore, $PQparallel BC$.
answered Nov 16 at 10:15
Batominovski
31.6k23188
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
|
show 5 more comments
up vote
4
down vote
up vote
4
down vote
By the Angle Bisector Theorem, $$frac{AQ}{QB}=frac{AM}{MB}text{ and }frac{AP}{PC}=frac{AM}{MC},.$$
Since $M$ is the midpoint of $BC$, we have $MB=MC$, whence
$$frac{AQ}{QB}=frac{AP}{PC},.$$
Therefore, $PQparallel BC$.
answered Nov 16 at 10:15
Batominovski
31.6k23188
By the Angle Bisector Theorem, $$frac{AQ}{QB}=frac{AM}{MB}text{ and }frac{AP}{PC}=frac{AM}{MC},.$$
Since $M$ is the midpoint of $BC$, we have $MB=MC$, whence
$$frac{AQ}{QB}=frac{AP}{PC},.$$
Therefore, $PQparallel BC$.
answered Nov 16 at 10:15
Batominovski
31.6k23188
edited Nov 16 at 10:21
answered Nov 16 at 10:15
Batominovski
31.6k23188
answered Nov 16 at 10:15
Batominovski
31.6k23188
answered Nov 16 at 10:15
Batominovski
31.6k23188
31.6k23188
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
|
show 5 more comments
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
This might be a silly question but how did you reach the last conclusion? I mean from $$frac{AQ}{QB} = frac{AP}{PC}$$ to $PQ || BC$
– Sauhard Sharma
Nov 16 at 10:22
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html.
– Batominovski
Nov 16 at 10:23
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram!
– user602338
Nov 16 at 10:46
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
@user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$.
– Batominovski
Nov 16 at 10:57
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
You have written PC in your proof above. Don't you?
– user602338
Nov 16 at 10:59
|
show 5 more comments
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000966%2fprove-that-in-triangle-abc-qp-parallel-bc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect.
– Eevee Trainer
Nov 16 at 10:13