Can't figure out this triangle geometry problem












5












$begingroup$


I have the following triangle:



enter image description here




The following information about it are given:




  • ABCD is a trapezoid (AB || DC)

  • EF || DC

  • Q is the intersection of AC, DB, PN, & EF


Prove that EQ = QF.




Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:



$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$



$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$



I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.



How can I solve this particular problem, and how do I tackle problems of this kind more effectively?






geometry euclidean-geometry geometric-transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
    $endgroup$
    – Matteo
    Jan 2 at 20:46












  • $begingroup$
    @Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
    $endgroup$
    – greedoid
    Jan 2 at 20:51










  • $begingroup$
    Q is just a point. No information about middle points is given.
    $endgroup$
    – daedsidog
    Jan 2 at 20:51
















5












$begingroup$


I have the following triangle:



enter image description here




The following information about it are given:




  • ABCD is a trapezoid (AB || DC)

  • EF || DC

  • Q is the intersection of AC, DB, PN, & EF


Prove that EQ = QF.




Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:



$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$



$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$



I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.



How can I solve this particular problem, and how do I tackle problems of this kind more effectively?






geometry euclidean-geometry geometric-transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
    $endgroup$
    – Matteo
    Jan 2 at 20:46












  • $begingroup$
    @Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
    $endgroup$
    – greedoid
    Jan 2 at 20:51










  • $begingroup$
    Q is just a point. No information about middle points is given.
    $endgroup$
    – daedsidog
    Jan 2 at 20:51














5












5








5





$begingroup$


I have the following triangle:



enter image description here




The following information about it are given:




  • ABCD is a trapezoid (AB || DC)

  • EF || DC

  • Q is the intersection of AC, DB, PN, & EF


Prove that EQ = QF.




Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:



$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$



$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$



I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.



How can I solve this particular problem, and how do I tackle problems of this kind more effectively?






geometry euclidean-geometry geometric-transformation







share|cite|improve this question











$endgroup$




I have the following triangle:



enter image description here




The following information about it are given:




  • ABCD is a trapezoid (AB || DC)

  • EF || DC

  • Q is the intersection of AC, DB, PN, & EF


Prove that EQ = QF.




Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:



$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$



$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$



I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.



How can I solve this particular problem, and how do I tackle problems of this kind more effectively?






geometry euclidean-geometry geometric-transformation




geometry euclidean-geometry geometric-transformation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 12:59









greedoid

46.1k1160117




46.1k1160117










asked Jan 2 at 20:36









daedsidogdaedsidog

29517




29517












  • $begingroup$
    It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
    $endgroup$
    – Matteo
    Jan 2 at 20:46












  • $begingroup$
    @Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
    $endgroup$
    – greedoid
    Jan 2 at 20:51










  • $begingroup$
    Q is just a point. No information about middle points is given.
    $endgroup$
    – daedsidog
    Jan 2 at 20:51


















  • $begingroup$
    It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
    $endgroup$
    – Matteo
    Jan 2 at 20:46












  • $begingroup$
    @Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
    $endgroup$
    – greedoid
    Jan 2 at 20:51










  • $begingroup$
    Q is just a point. No information about middle points is given.
    $endgroup$
    – daedsidog
    Jan 2 at 20:51
















$begingroup$
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
$endgroup$
– Matteo
Jan 2 at 20:46






$begingroup$
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
$endgroup$
– Matteo
Jan 2 at 20:46














$begingroup$
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
$endgroup$
– greedoid
Jan 2 at 20:51




$begingroup$
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
$endgroup$
– greedoid
Jan 2 at 20:51












$begingroup$
Q is just a point. No information about middle points is given.
$endgroup$
– daedsidog
Jan 2 at 20:51




$begingroup$
Q is just a point. No information about middle points is given.
$endgroup$
– daedsidog
Jan 2 at 20:51










3 Answers
3






active

oldest

votes


















2












$begingroup$

First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).



$$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
    $endgroup$
    – daedsidog
    Jan 2 at 21:01



















0












$begingroup$

Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.



Let’s first prove that $P, M, Q, N$ are aligned.



The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.



Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.



Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    By similarity twice and by Thales for $angle DPC$ we obtain:
    $$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).



      $$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
        $endgroup$
        – daedsidog
        Jan 2 at 21:01
















      2












      $begingroup$

      First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).



      $$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
        $endgroup$
        – daedsidog
        Jan 2 at 21:01














      2












      2








      2





      $begingroup$

      First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).



      $$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$






      share|cite|improve this answer









      $endgroup$



      First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).



      $$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 2 at 20:56









      greedoidgreedoid

      46.1k1160117




      46.1k1160117












      • $begingroup$
        Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
        $endgroup$
        – daedsidog
        Jan 2 at 21:01


















      • $begingroup$
        Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
        $endgroup$
        – daedsidog
        Jan 2 at 21:01
















      $begingroup$
      Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
      $endgroup$
      – daedsidog
      Jan 2 at 21:01




      $begingroup$
      Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
      $endgroup$
      – daedsidog
      Jan 2 at 21:01











      0












      $begingroup$

      Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.



      Let’s first prove that $P, M, Q, N$ are aligned.



      The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.



      Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.



      Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.



        Let’s first prove that $P, M, Q, N$ are aligned.



        The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.



        Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.



        Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.



          Let’s first prove that $P, M, Q, N$ are aligned.



          The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.



          Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.



          Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.






          share|cite|improve this answer









          $endgroup$



          Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.



          Let’s first prove that $P, M, Q, N$ are aligned.



          The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.



          Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.



          Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 21:03









          mathcounterexamples.netmathcounterexamples.net

          27k22158




          27k22158























              0












              $begingroup$

              By similarity twice and by Thales for $angle DPC$ we obtain:
              $$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                By similarity twice and by Thales for $angle DPC$ we obtain:
                $$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  By similarity twice and by Thales for $angle DPC$ we obtain:
                  $$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$






                  share|cite|improve this answer









                  $endgroup$



                  By similarity twice and by Thales for $angle DPC$ we obtain:
                  $$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 22:01









                  Michael RozenbergMichael Rozenberg

                  107k1894199




                  107k1894199






























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