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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.
Solution
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.
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