558 
NOTE ON THE TETRAHEDRON. 
[382 
is at once seen to be 
2 («i - Oj) («i + a 2 - «s - <* 4 ) = 0, 
where 2 denotes the sum of the corresponding terms in a, /3, y. And so the 
condition that the same line may cut at right angles the line CD is 
2 (a 3 - a 4 ) (at: + a 2 - a 3 - a 4 ) = 0. 
But the conditions AC = BD, AD = BC give respectively 
2 {(ai - a 3 ) 2 - (a 2 - a 4 ) 2 } = 0, 2 {(ax - a 4 ) 2 - (a 2 - a 3 ) 2 } = 0, 
or writing these in the form 
2 (<*! - a, -a s + <z 4 ) (a x + a, - a 3 - a 4 ) = 0, 
2 (a: - a 2 + a 3 - a 4 ) (a! + a, - a 3 - a 4 ) = 0, 
we obtain, by successively adding and subtracting, the two required equations. 
The equations of the line through £AC, %BD are 
a ~ 2 ( a i + *») = y-b(0i + &) = * ~ i (71 + 7») 
«i + a 3 - a, - a 4 &+&-&- & 7i + y 3 - 7, - 74 ’ 
and those of the line through \ AD, ^ BC are 
+ °0 = y- H& + A) = A ~ 2 (7i + 74) . 
a, + a 3 - acj - a 4 & + & - ft - & 72 + 7s ~ 7i - y 4 ’ 
and the condition that these may cut at right angles is 
2 (a, + a 3 - * 2 - a*) (a! + a 4 - a 2 - a 3 ) = 0, 
that is 
2 {(<*1 - a 2 ) 2 - (>3 - 0O 2 } = 0, 
which is in fact the condition AB = CD. 
Combining the two theorems we see that if in a tetrahedron the pairs of opposite 
sides are respectively equal, then the line joining the centres of opposite sides cuts 
these sides at right angles, and moreover the three joining lines cut each other at 
right angles. 
A tetrahedron of the form in question may be constructed as follows: viz. taking 
a parallelogram ABCD, whereof the diagonals AC, BD are unequal, then bending the 
parallelogram about its shorter diagonal AC in such manner that in the solid figure 
BD becomes equal to AC, we have a tetrahedron the opposite sides whereof are 
respectively equal.