Q.If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular, then show that the edges in the third pair are also perpendicular.

Write each midpoint's position vector using the midpoint formula, then add the three vectors.

Let position vectors of $A,B,C$ be $\bar a,\bar b,\bar c$ (origin $O$). Since $D,E,F$ are midpoints of $BC,CA,AB$:

$\bar d=\frac{\bar b+\bar c}{2},\qquad \bar e=\frac{\bar c+\bar a}{2},\qquad \bar f=\frac{\bar a+\bar b}{2}$

$\overline{AD}=\bar d-\bar a=\frac{\bar b+\bar c}{2}-\bar a$

$\overline{BE}=\bar e-\bar b=\frac{\bar c+\bar a}{2}-\bar b$

$\overline{CF}=\bar f-\bar c=\frac{\bar a+\bar b}{2}-\bar c$

Adding: …

Use $A=-B$ (diametrically opposite about center $O$) and show $\vec{PA}\cdot\vec{PB}=0$.

Let $O$ be the centre, radius $r$, with $A,B$ the endpoints of a diameter, so $\vec{OA}=-\vec{OB}$ (i.e. $\bar a=-\bar b$). Let $P$ be any point on the circle, $\vec{OP}=\bar p$, with $|\bar p|=|\bar a|=|\bar b|=r$.

$\vec{PA}=\bar a-\bar p=-\bar b-\bar p$

$\vec{PB}=\bar b-\bar p$

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Show the point dividing each median in ratio $2:1$ from the vertex is the same point $\dfrac{\bar a+\bar b+\bar c}{3}$ for all three medians.

Let $\triangle ABC$ have position vectors $\bar a, \bar b, \bar c$ with respect to origin $O$.

Let $D$ be the midpoint of $BC$. Then

$\bar d = \frac{\bar b + \bar c}{2}$

Let $G$ be the point on median $AD$ dividing it internally in the ratio $2:1$ from $A$ (i.e. $AG:GD = 2:1$). By the section formula:

$\bar g = \frac{2\bar d + 1\cdot\bar a}{2+1} = \frac{2\left(\dfrac{\bar b+\bar c}{2}\right)+\bar a}{3} = \frac{\bar a+\bar b+\bar c}{3}$

Now let $E$ be the midpoint of $AC$, so $\bar e = \dfrac{\bar a+\bar c}{2}$. Let $G'$ divide median $BE$ in ratio $2:1$ from $B$:

$\bar{g'} = \frac{2\bar e + \bar b}{3} = \frac{2\left(\dfrac{\bar a+\bar c}{2}\right)+\bar b}{3} = \frac{\bar a+\bar b+\bar c}{3}$

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Show the point dividing each median in ratio $2:1$ from the vertex is the same point $\dfrac{\bar a+\bar b+\bar c}{3}$ for all three medians.

Let $\triangle ABC$ have position vectors $\bar a,\bar b,\bar c$ for its vertices A, B, C respectively.

Median from A: Let D be the midpoint of BC, so $\bar d=\dfrac{\bar b+\bar c}{2}$. Let G be the point on AD dividing it internally in ratio $2:1$ (from A):

$\bar g=\frac{2\bar d+1\cdot\bar a}{2+1}=\frac{2\left(\frac{\bar b+\bar c}{2}\right)+\bar a}{3}=\frac{\bar a+\bar b+\bar c}{3}$

Median from B: Let E be the midpoint of AC, $\bar e=\dfrac{\bar a+\bar c}{2}$. The point dividing BE in ratio $2:1$ from B is:

$\frac{2\bar e+\bar b}{3}=\frac{2\left(\frac{\bar a+\bar c}{2}\right)+\bar b}{3}=\frac{\bar a+\bar b+\bar c}{3}$

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