`A B C D` is a tetrahedron and `O` is any point. If the lines joining `O` to the vrticfes meet the opposite faces at `P ,Q ,Ra n dS ,` prove that `(O P)/(A P)+(O Q)/(B Q)+(O R)/(C R)+(O S)/(D S)=1.`

Explanation

Here ABCD is a tetrahedron. Let O be the origin and the P.V of A, B, C and Dd be `veca, vecb, vecc and vecd`, respectively.
We know that four linearly dependent vectors can be expressed as
`" " xveca+ yvecb+zvecc+ tvecd=0`
(where x, y, z and t are scalars)
or `" "yvecb+zvecc+ tvecd= - xveca`
or `" "= (yvecb+zvecc+tvecd)/(y+z+t) = - (xveca)/(y+z+t)`
where L.H.S is P.V of a point in the palne BCD and R.H.S is a point on `vec(AO)` .
Therefore, there must be a point common to both the plane and the straight line. That is
`" "vec(OP) = (-xveca)/(y+z+t)`
But, `vec(AP)=vec(OP) -vec(OA) = - (xveca)/( y+z+t)-veca`
`" "=-((x+y+z+t)/(y+z+t))veca`
`vec(OP) = (x) /(y+z+t) ((y+z+t)/(x+y+z+t))vec(AP)`
`" " = ((x)/(x+y+z+t))vec(AP)`
`rArr (OP)/(AP) = (x)/(x+y+z+t)`
Similarly, `(OQ)/(BQ) = (y)/(x+y+z+t)`
`" "(OR)/(CR) = (z)/(x+y+z+t) and (OS)/(DS) = (t)/(x+y+z+t)`
`rArr (OP)/(AP) + (OQ)/(BQ) + (OR)/(CR) + (OS)/(DS) = 1 `