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The concurrency of the heights of a tetrahedron [duplicate]

Concurrency of the heights of a tetrahedron with opposite edges perpendicular.What determines the height of a tetrahedron?Center of mass of a tetrahedronProve a specific property for tetrahedronConcurrency ProofSolve a tetrahedron with given lengths of three edges and a certain propertyConcurrency of the heights of a tetrahedron with opposite edges perpendicular.Find all the properties of tetrahedron if the lengths of all the edges are known.Circumradius of a tetrahedronThe midpoints of an isosceles tetrahedronA problem with tetrahedron

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This question already has an answer here:

• 
  Concurrency of the heights of a tetrahedron with opposite edges perpendicular.
  
  2 answers
  
  

The sentence says that the opposite edges of the tetrahedron ABCD are perpendicular and the fact that $AB^2+CD^2=AC^2+BD^2=BC^2+AD^2$ and I need to prove that the heights of the tetrahedron are concurrent, but I am stuck. I need a vectorial proof.The picture of my solution so far

geometry analytic-geometry

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edited Mar 28 at 20:25

MPW

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asked Mar 28 at 20:23

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-1

$begingroup$

This question already has an answer here:

• 
  Concurrency of the heights of a tetrahedron with opposite edges perpendicular.
  
  2 answers
  
  

The sentence says that the opposite edges of the tetrahedron ABCD are perpendicular and the fact that $AB^2+CD^2=AC^2+BD^2=BC^2+AD^2$ and I need to prove that the heights of the tetrahedron are concurrent, but I am stuck. I need a vectorial proof.The picture of my solution so far

geometry analytic-geometry

share|cite|improve this question

edited Mar 28 at 20:25

MPW

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asked Mar 28 at 20:23

SR7SR7

1

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Mar 28 at 21:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$begingroup$

This question already has an answer here:

• 
  Concurrency of the heights of a tetrahedron with opposite edges perpendicular.
  
  2 answers
  
  

The sentence says that the opposite edges of the tetrahedron ABCD are perpendicular and the fact that $AB^2+CD^2=AC^2+BD^2=BC^2+AD^2$ and I need to prove that the heights of the tetrahedron are concurrent, but I am stuck. I need a vectorial proof.The picture of my solution so far

geometry analytic-geometry

share|cite|improve this question

edited Mar 28 at 20:25

MPW

31k12157

asked Mar 28 at 20:23

SR7SR7

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edited Mar 28 at 20:25

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$begingroup$

Let $vecDA=veca,$ $vecDB=vecb$ and $vecDC=vecc.$

Thus, $$AB^2+CD^2=AC^2+BD^2$$ it's
$$(vecb-veca)^2+vecc^2=(vecc-veca)^2+vecb^2$$ or
$$vecavecb=vecavecc$$ or
$$veca(vecb-vecc)=0$$ or $$ADperp BC.$$
Similarly, we obtain that $BDperp AC$ and $CDperp AB.$

Now, let $DK$ be an altitude of the tetrahedron and $AKcap BC=E.$

Also, let $AF$ be an altitude of $Delta ADE$.

We see that $BCperp AE$ and $BCperp DK$.

Thus, $BCperp(ADE)$ and since $AFsubset(ADC),$ we obtain $BCperp AF$.

Id est, $AFperp BC$ and $AFperp DF,$ which says $AFperp(DBC)$ and we got that $AF$ is an altitude of the tetrahedron.

But $AFsubset(ADE)$ and $DKsubset(ADE),$ which says that $AF$ and $DK$ intersect.

Can you end it now?

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edited Mar 28 at 21:42

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

add a comment | 

0

$begingroup$

Let $vecDA=veca,$ $vecDB=vecb$ and $vecDC=vecc.$

Thus, $$AB^2+CD^2=AC^2+BD^2$$ it's
$$(vecb-veca)^2+vecc^2=(vecc-veca)^2+vecb^2$$ or
$$vecavecb=vecavecc$$ or
$$veca(vecb-vecc)=0$$ or $$ADperp BC.$$
Similarly, we obtain that $BDperp AC$ and $CDperp AB.$

Now, let $DK$ be an altitude of the tetrahedron and $AKcap BC=E.$

Also, let $AF$ be an altitude of $Delta ADE$.

We see that $BCperp AE$ and $BCperp DK$.

Thus, $BCperp(ADE)$ and since $AFsubset(ADC),$ we obtain $BCperp AF$.

Id est, $AFperp BC$ and $AFperp DF,$ which says $AFperp(DBC)$ and we got that $AF$ is an altitude of the tetrahedron.

But $AFsubset(ADE)$ and $DKsubset(ADE),$ which says that $AF$ and $DK$ intersect.

Can you end it now?

share|cite|improve this answer

edited Mar 28 at 21:42

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

add a comment | 

0

$begingroup$

Let $vecDA=veca,$ $vecDB=vecb$ and $vecDC=vecc.$

Thus, $$AB^2+CD^2=AC^2+BD^2$$ it's
$$(vecb-veca)^2+vecc^2=(vecc-veca)^2+vecb^2$$ or
$$vecavecb=vecavecc$$ or
$$veca(vecb-vecc)=0$$ or $$ADperp BC.$$
Similarly, we obtain that $BDperp AC$ and $CDperp AB.$

Now, let $DK$ be an altitude of the tetrahedron and $AKcap BC=E.$

Also, let $AF$ be an altitude of $Delta ADE$.

We see that $BCperp AE$ and $BCperp DK$.

Thus, $BCperp(ADE)$ and since $AFsubset(ADC),$ we obtain $BCperp AF$.

Id est, $AFperp BC$ and $AFperp DF,$ which says $AFperp(DBC)$ and we got that $AF$ is an altitude of the tetrahedron.

But $AFsubset(ADE)$ and $DKsubset(ADE),$ which says that $AF$ and $DK$ intersect.

Can you end it now?

share|cite|improve this answer

edited Mar 28 at 21:42

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

add a comment | 

$begingroup$

Let $vecDA=veca,$ $vecDB=vecb$ and $vecDC=vecc.$

Thus, $$AB^2+CD^2=AC^2+BD^2$$ it's
$$(vecb-veca)^2+vecc^2=(vecc-veca)^2+vecb^2$$ or
$$vecavecb=vecavecc$$ or
$$veca(vecb-vecc)=0$$ or $$ADperp BC.$$
Similarly, we obtain that $BDperp AC$ and $CDperp AB.$

Now, let $DK$ be an altitude of the tetrahedron and $AKcap BC=E.$

Also, let $AF$ be an altitude of $Delta ADE$.

We see that $BCperp AE$ and $BCperp DK$.

Thus, $BCperp(ADE)$ and since $AFsubset(ADC),$ we obtain $BCperp AF$.

Id est, $AFperp BC$ and $AFperp DF,$ which says $AFperp(DBC)$ and we got that $AF$ is an altitude of the tetrahedron.

But $AFsubset(ADE)$ and $DKsubset(ADE),$ which says that $AF$ and $DK$ intersect.

Can you end it now?

share|cite|improve this answer

edited Mar 28 at 21:42

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

share|cite|improve this answer

edited Mar 28 at 21:42

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201

answered Mar 28 at 21:34

Michael RozenbergMichael Rozenberg

109k1896201