Prove that a line is a bisector of the angle












2












$begingroup$


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 $D$?enter image description here






geometry euclidean-geometry







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  • $begingroup$
    Are you interested in a proof using vectors?
    $endgroup$
    – David Quinn
    May 21 '17 at 19:20
















2












$begingroup$


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 $D$?enter image description here






geometry euclidean-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you interested in a proof using vectors?
    $endgroup$
    – David Quinn
    May 21 '17 at 19:20














2












2








2


2



$begingroup$


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 $D$?enter image description here






geometry euclidean-geometry







share|cite|improve this question











$endgroup$




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 $D$?enter image description here






geometry euclidean-geometry




geometry euclidean-geometry






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edited May 21 '17 at 18:18







idliketodothis

















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


















  • $begingroup$
    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










1 Answer
1






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2












$begingroup$

enter image description here



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 $Delta ADP$.






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  • $begingroup$
    @idliketodothis: Is it clear?
    $endgroup$
    – Arnaldo
    May 22 '17 at 17:25











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1 Answer
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1 Answer
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2












$begingroup$

enter image description here



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 $Delta ADP$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @idliketodothis: Is it clear?
    $endgroup$
    – Arnaldo
    May 22 '17 at 17:25
















2












$begingroup$

enter image description here



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 $Delta ADP$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @idliketodothis: Is it clear?
    $endgroup$
    – Arnaldo
    May 22 '17 at 17:25














2












2








2





$begingroup$

enter image description here



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 $Delta ADP$.






share|cite|improve this answer









$endgroup$



enter image description here



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 $Delta ADP$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    @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


















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