Prove that a line is a bisector of the angle
$begingroup$
In a parallelogram $ABCD$ points $M$ and $N$ were chosen on the sides $AB$ and $BC$ respectively. $AM = NC$. The point $Q$ is the intersection point of the line segments $AN$ and $CM$. How can I prove that $DQ$ is the bisector of the angle $D$?
geometry euclidean-geometry
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
$endgroup$
add a comment |
$begingroup$
In a parallelogram $ABCD$ points $M$ and $N$ were chosen on the sides $AB$ and $BC$ respectively. $AM = NC$. The point $Q$ is the intersection point of the line segments $AN$ and $CM$. How can I prove that $DQ$ is the bisector of the angle $D$?
geometry euclidean-geometry
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
$endgroup$
$begingroup$
Are you interested in a proof using vectors?
$endgroup$
– David Quinn
May 21 '17 at 19:20
add a comment |
$begingroup$
In a parallelogram $ABCD$ points $M$ and $N$ were chosen on the sides $AB$ and $BC$ respectively. $AM = NC$. The point $Q$ is the intersection point of the line segments $AN$ and $CM$. How can I prove that $DQ$ is the bisector of the angle $D$?
geometry euclidean-geometry
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
$endgroup$
In a parallelogram $ABCD$ points $M$ and $N$ were chosen on the sides $AB$ and $BC$ respectively. $AM = NC$. The point $Q$ is the intersection point of the line segments $AN$ and $CM$. How can I prove that $DQ$ is the bisector of the angle $D$?
geometry euclidean-geometry
geometry euclidean-geometry
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
edited May 21 '17 at 18:18
idliketodothis
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
asked May 21 '17 at 18:11
idliketodothisidliketodothis
45926
45926
$begingroup$
Are you interested in a proof using vectors?
$endgroup$
– David Quinn
May 21 '17 at 19:20
add a comment |
$begingroup$
Are you interested in a proof using vectors?
$endgroup$
– David Quinn
May 21 '17 at 19:20
$begingroup$
Are you interested in a proof using vectors?
$endgroup$
– David Quinn
May 21 '17 at 19:20
$begingroup$
Are you interested in a proof using vectors?
$endgroup$
– David Quinn
May 21 '17 at 19:20
add a comment |
1 Answer
1
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votes
$begingroup$
Let's call $P$ the intersection between the lines through $AN$ and $DC$.
So, $Delta QCP$ and $Delta MQA$ are similars, and then
$$frac{QP}{AQ}=frac{PC}{AM}quad (1)$$
we also have $Delta NPC$ and $Delta ABN$ are similars, then
$$frac{PC}{AB}=frac{NC}{BN}tofrac{PC}{AB}=frac{AM}{BN}to frac{PC}{AM}=frac{AB}{BN}quad (2)$$
but $Delta ABN$ and $Delta ADP$ are also similars, then
$$frac{AB}{PD}=frac{BN}{AD}to frac{AB}{BN}=frac{PD}{AD}quad (3)$$
putting together $(1), (2), (3)$ we get:
$$frac{QP}{AQ}=frac{PD}{AD}$$
and by bissector theorem we get that $QD$ is the bissector of the angle $angle D$ in the triangle $Delta ADP$.
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
$endgroup$
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's call $P$ the intersection between the lines through $AN$ and $DC$.
So, $Delta QCP$ and $Delta MQA$ are similars, and then
$$frac{QP}{AQ}=frac{PC}{AM}quad (1)$$
we also have $Delta NPC$ and $Delta ABN$ are similars, then
$$frac{PC}{AB}=frac{NC}{BN}tofrac{PC}{AB}=frac{AM}{BN}to frac{PC}{AM}=frac{AB}{BN}quad (2)$$
but $Delta ABN$ and $Delta ADP$ are also similars, then
$$frac{AB}{PD}=frac{BN}{AD}to frac{AB}{BN}=frac{PD}{AD}quad (3)$$
putting together $(1), (2), (3)$ we get:
$$frac{QP}{AQ}=frac{PD}{AD}$$
and by bissector theorem we get that $QD$ is the bissector of the angle $angle D$ in the triangle $Delta ADP$.
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
$endgroup$
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
add a comment |
$begingroup$
Let's call $P$ the intersection between the lines through $AN$ and $DC$.
So, $Delta QCP$ and $Delta MQA$ are similars, and then
$$frac{QP}{AQ}=frac{PC}{AM}quad (1)$$
we also have $Delta NPC$ and $Delta ABN$ are similars, then
$$frac{PC}{AB}=frac{NC}{BN}tofrac{PC}{AB}=frac{AM}{BN}to frac{PC}{AM}=frac{AB}{BN}quad (2)$$
but $Delta ABN$ and $Delta ADP$ are also similars, then
$$frac{AB}{PD}=frac{BN}{AD}to frac{AB}{BN}=frac{PD}{AD}quad (3)$$
putting together $(1), (2), (3)$ we get:
$$frac{QP}{AQ}=frac{PD}{AD}$$
and by bissector theorem we get that $QD$ is the bissector of the angle $angle D$ in the triangle $Delta ADP$.
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
$endgroup$
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
add a comment |
$begingroup$
Let's call $P$ the intersection between the lines through $AN$ and $DC$.
So, $Delta QCP$ and $Delta MQA$ are similars, and then
$$frac{QP}{AQ}=frac{PC}{AM}quad (1)$$
we also have $Delta NPC$ and $Delta ABN$ are similars, then
$$frac{PC}{AB}=frac{NC}{BN}tofrac{PC}{AB}=frac{AM}{BN}to frac{PC}{AM}=frac{AB}{BN}quad (2)$$
but $Delta ABN$ and $Delta ADP$ are also similars, then
$$frac{AB}{PD}=frac{BN}{AD}to frac{AB}{BN}=frac{PD}{AD}quad (3)$$
putting together $(1), (2), (3)$ we get:
$$frac{QP}{AQ}=frac{PD}{AD}$$
and by bissector theorem we get that $QD$ is the bissector of the angle $angle D$ in the triangle $Delta ADP$.
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
$endgroup$
Let's call $P$ the intersection between the lines through $AN$ and $DC$.
So, $Delta QCP$ and $Delta MQA$ are similars, and then
$$frac{QP}{AQ}=frac{PC}{AM}quad (1)$$
we also have $Delta NPC$ and $Delta ABN$ are similars, then
$$frac{PC}{AB}=frac{NC}{BN}tofrac{PC}{AB}=frac{AM}{BN}to frac{PC}{AM}=frac{AB}{BN}quad (2)$$
but $Delta ABN$ and $Delta ADP$ are also similars, then
$$frac{AB}{PD}=frac{BN}{AD}to frac{AB}{BN}=frac{PD}{AD}quad (3)$$
putting together $(1), (2), (3)$ we get:
$$frac{QP}{AQ}=frac{PD}{AD}$$
and by bissector theorem we get that $QD$ is the bissector of the angle $angle D$ in the triangle $Delta ADP$.
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
answered May 21 '17 at 18:57
ArnaldoArnaldo
18.2k42246
18.2k42246
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
add a comment |
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
$begingroup$
@idliketodothis: Is it clear?
$endgroup$
– Arnaldo
May 22 '17 at 17:25
add a comment |
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$begingroup$
Are you interested in a proof using vectors?
$endgroup$
– David Quinn
May 21 '17 at 19:20