Question

ABCD is a parallelogram. E, F are the midpoints 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 `bar"a",bar"b",bar"c",bar"d",bar"e",bar"f",bar"p",bar"q"` respectively.

∵ ABCD is a parallelogram

∴ `bar"AB" = bar"DC"`

∴ `bar"b" - bar"a" = bar"c" - bar"d"`

∴ `bar"c" = bar"b" + bar"d" - bar"a"`        ....(1)

E is the midpoint of BC

∴ `bar"e" = (bar"b" + bar"c")/2`

∴ `2bar"e" = bar"b" + bar"c"`     ....(2)

F is the mid-point of CD

∴ `bar"f" = (bar"c" + bar"d")/2`

∴ `2bar"f" = bar"c" + bar"d"`        .....(3)

`2bar"e" = bar"b" + bar"c"`      ...[By (2)]

`= bar"b" + (bar"b" + bar"d" + bar"a")`    ....[By (1)]

∴ `2bar"e" + bar"a" = 2bar"b" + bar"d"`

∴ `(2bar"e" + bar"a")/(2+1) = (2bar"b" + bar"d")/(2 + 1)`

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.

∴ `bar"q" = (2bar"b" + bar"d")/(2 + 1)`

∴ Q divides DB in the ratio 2 : 1      ....(4)

`2bar"f" = bar"c" + bar"d"`      ....[By(3)]

`= (bar"b" + bar"d" - bar"a") + bar"d"`      ....[By(1)]

∴ `2bar"f" + bar"a" = 2bar"d" + bar"b"`

∴ `(bar"a" + 2bar"f")/(1 + 2) = (bar"b" + 2bar"d")/(1 + 2)`

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.

∴ `bar"p" = (bar"b" + 2bar"d")/(1 + 2)`

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

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