Question

ABCD is a parallelogram E, F are the mid points of BC and CD respectively. AE, AF meet the diagonal BD at Q and P respectively. Show that P and Q trisect DB.

Answer 1

Let A, B, C, D, E, F, P, Q have position vectors \(\overline{a},\) \(\overline{b},\) \(\overline{c},\) \(\overline{d},\) \(\overline{e},\) \(\overline{f},\) \(\overline{p},\) \(\overline{q}\) respectively.

\(\because\) ABCD is a parallelogram

LHS is the position vector of the point on AE and RHS is the position vector of the point on DB. But AE and DB meet at Q.

LHS is the position vector of the point on AF and RHS is the position vector of the point on DB.

But AF and DB meet at P.

∴ P divides DB in the ratio 1 : 2 … (5) 

From (4) and (5), if follows that P and Q trisect DB.