Q.286. Property of a regular tetrahedron.

Question 286.
Let O he any point within a regular tetrahedron ABCD. Lines joining the vertices A, B, C, D to O when produced meet the opposite faces in P, Q, R and S respectively. Prove that OP/AP + OQ/BQ +OR/CR+ OS/DS = 1.

Answer 286.
Join O to A, B, C and D.

It divides the tetrahedron ABCD into four smaller tetrahedrons, OBCD, OCDA, ODAB and OABC.

Let x = area of traingular base of tetrahedron and h = altitude from any vertex to the opposite face.
Let P', Q', R' and S' be the feet of perpendiculars from O on faces opposite to A, B, C and D.
=> OP/AP = OP'/h, OQ/BQ = OQ'/h, OR/CR = OR'/h and OS/DS = OS'/h

Sum of volumes of the four smaller tetrahedrons, OBCD, OCDA, ODAB and OABC
= volume of the larger tetrahedron ABCD.
=> (1/6) [OP' *x + OQ' * x + OR' * x + OS' * x) = (1/6) h * x
=> x [h * OP/AP + h * OQ/BQ + h * OR/CR + h * OS/DS] = x * h
=> OP/AP + OQ/BQ +OR/CR+ OS/DS = 1.

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