In space, four points, $A$, $B$, $C$, and $D$ are given such that $AB = AC$ and $DB = DC$. Prove that the lines $AD$ and $BC$ are perpendicular.

Question

Exercise

In space, four points, $A$, $B$, $C$, and $D$ are given such that $AB = AC$ and $DB = DC$. Prove that the lines $AD$ and $BC$ are perpendicular.

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Attempt

My attempt included drawing a plane $M$ parallel to $AD$ and $BC$ (by first drawing a lines $l$ perpendicular to $AD$ and $BC$ (or their extensions)).

If I can prove that the projection of $ABCD$ is a diamond, then I'll bet fine, because I've already proved that the diagonals of diamonds are perpendicular to each other.

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Postscript

This is exercise 43 of Kiselev's Geometry: Book 2: Stereometry.

Answer 1

In typing up this question and providing my attempt, I lightbulb appeared over my head. :)

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Proof

1. Draw line $l$ such that $l \perp AC$ and $l \perp BD$.
2. Draw plane $M$ such that $M \perp l$.
3. Drop perpendiculars $AA'$, $BB'$, $CC'$, and $DD'$ onto $M$.
4. $AA'D'D = CC'D'D$ and $AA'B'B = CC'B'B$
5. $A'D' = C'D'$ and $A'B' = C'B'$
6. $A'B'C'D'$ is a diamond
7. $A'C' \perp B'D'$

$\square$

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Answer 2

Here is a detailed reasoning of why this is true. I suggest, draw a picture on a piece of paper.

The four points form a tetrahedron, labeled $ABCD$. Take the midpoint $M$ of segment $BC$. Draw the lines $AM$ and $DM$ passing though the point $M$. Two lines, passing through a common point span a plane, in this case it is the plane $ADM$. Notice that $AD$ lies on the plane $ADM$ because both points $A$ and $D$ lie in it.

Notice that in triangle $ABC$ segment $AM$ is a median and since $AB = AC$, the triangle $ABC$ is isosceles so $AM$ is a orthogonal bisector of edge $BC$, i.e. $BM$ is orthogonal to $AM$. Analogously, in triangle $DBC$ segment $DM$ is a median and since $DB = DC$, the triangle $DBC$ is isosceles so $DM$ is a orthogonal bisector of edge $BC$, i.e. $BC$ is orthogonal to $DM$.

Since, $BC$ is orthogonal to two transverse lines, $AM$ and $DM$, in the plane $ADM$, the line $BC$ is orthogonal to the whole plane $ADM$ and thus it is orthogonal to all lines on the plane $ADM$. In particular $BC$ is orthogonal to $AD$.