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proof for trapezoid within triangle conundrum

Proving that two crossbars of a bisector intersect the midpoint of one of the crossbarsTrapezoid and isosceles triangleProving a quadrilateral is an isosceles trapezoidTriangle's Area within a TrapezoidTriangle - Trapezoid [Geometry]Right-angled triangle proofTrapezoid midsegment diagonal proofInscribing rhombus in a triangle's angle in only eight compass-and-straightedge stepsEquilateral Triangle within a RectangleTriangle equilateral proof

2

$begingroup$

I've recently been working on some geometry problems involving non-isosceles trapezoids and stumbled upon an interesting discovery which seems to hold true for all cases I've tried, but I don't have the expertise to construct a proof.
Here's the thing:

1. take any random triangle ABC


2. choose two points Q on edge AB and P on edge AC such that QP || BC. We now have constructed a trapezoid QBCP



3. find K the intersection point of the trapezoid's diagonals.




4. The line AK will perfectly bisect QP and BC in the middle. This is what requires proof.

I am very interested if somebody can construct a proof for this.
Thanks

illustration

geometry euclidean-geometry triangle

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

New contributor

Bogdan Ionitza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

I've recently been working on some geometry problems involving non-isosceles trapezoids and stumbled upon an interesting discovery which seems to hold true for all cases I've tried, but I don't have the expertise to construct a proof.
Here's the thing:

1. take any random triangle ABC


2. choose two points Q on edge AB and P on edge AC such that QP || BC. We now have constructed a trapezoid QBCP



3. find K the intersection point of the trapezoid's diagonals.




4. The line AK will perfectly bisect QP and BC in the middle. This is what requires proof.

I am very interested if somebody can construct a proof for this.
Thanks

illustration

geometry euclidean-geometry triangle

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

New contributor

Bogdan Ionitza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I've recently been working on some geometry problems involving non-isosceles trapezoids and stumbled upon an interesting discovery which seems to hold true for all cases I've tried, but I don't have the expertise to construct a proof.
Here's the thing:

1. take any random triangle ABC


2. choose two points Q on edge AB and P on edge AC such that QP || BC. We now have constructed a trapezoid QBCP



3. find K the intersection point of the trapezoid's diagonals.




4. The line AK will perfectly bisect QP and BC in the middle. This is what requires proof.

I am very interested if somebody can construct a proof for this.
Thanks

illustration

geometry euclidean-geometry triangle

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

New contributor

Bogdan Ionitza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

New contributor

Bogdan Ionitza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

New contributor

Bogdan Ionitza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

Michael Rozenberg

107k1895199

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

New contributor

Bogdan Ionitza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

Bogdan IonitzaBogdan Ionitza

132

1 Answer
1

active

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votes

0

$begingroup$

Let $AKcap QP=M$, $AKcap BC=N$, $QM=a$, $MP=b$, $BN=c$ and $NC=d$.

Thus, by similarity of triangles we obtain:
$$fracac=fracAMAN=fracbd$$ and
$$fracad=fracMKKN=fracbc,$$ which gives $a=b$ and $c=d$.

share|cite|improve this answer

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that seems correct. I didn't realize that the two pairs of opposing triangles defined by the trapezoid's diagonals, which share vertex K are also similar. That explains it.
  $endgroup$
  – Bogdan Ionitza
  2 days ago
  

  
  
  
  

add a comment | 

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0

$begingroup$

Let $AKcap QP=M$, $AKcap BC=N$, $QM=a$, $MP=b$, $BN=c$ and $NC=d$.

Thus, by similarity of triangles we obtain:
$$fracac=fracAMAN=fracbd$$ and
$$fracad=fracMKKN=fracbc,$$ which gives $a=b$ and $c=d$.

share|cite|improve this answer

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that seems correct. I didn't realize that the two pairs of opposing triangles defined by the trapezoid's diagonals, which share vertex K are also similar. That explains it.
  $endgroup$
  – Bogdan Ionitza
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $AKcap QP=M$, $AKcap BC=N$, $QM=a$, $MP=b$, $BN=c$ and $NC=d$.

Thus, by similarity of triangles we obtain:
$$fracac=fracAMAN=fracbd$$ and
$$fracad=fracMKKN=fracbc,$$ which gives $a=b$ and $c=d$.

share|cite|improve this answer

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that seems correct. I didn't realize that the two pairs of opposing triangles defined by the trapezoid's diagonals, which share vertex K are also similar. That explains it.
  $endgroup$
  – Bogdan Ionitza
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $AKcap QP=M$, $AKcap BC=N$, $QM=a$, $MP=b$, $BN=c$ and $NC=d$.

Thus, by similarity of triangles we obtain:
$$fracac=fracAMAN=fracbd$$ and
$$fracad=fracMKKN=fracbc,$$ which gives $a=b$ and $c=d$.

share|cite|improve this answer

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

share|cite|improve this answer

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199