asteroidfodder
New member
- Joined
- Jul 23, 2009
- Messages
- 6
The diagram shows a parallelogram OABC.
[attachment=0:31ao7upo]parallelogram.jpg[/attachment:31ao7upo]
The text uses notation of, for example, OA with an arrow above to indicate a vector. I will use O->A as a clumsy substitute.
O->A = b
O->C = c
AC and OB intersect at point D such that O->D = h(O->B) and A->D = k(A->C)
Show that h = k = 1/2.
I first thought to show that O->D = D->B.
O->B = a + c
so O->D = h(O->B) = h(a + c)
and similarly D->B = (1 - h) (a + c)
but
O->D + D->C = c and also A->D + D->B = c
which gives me
h(a+c) + (1-k)(c-a) = k(c-a) + (1-h) (a+c)
which is true but doesn't prove anything!
So I would be grateful for a hint. Thanks.
[attachment=0:31ao7upo]parallelogram.jpg[/attachment:31ao7upo]
The text uses notation of, for example, OA with an arrow above to indicate a vector. I will use O->A as a clumsy substitute.
O->A = b
O->C = c
AC and OB intersect at point D such that O->D = h(O->B) and A->D = k(A->C)
Show that h = k = 1/2.
I first thought to show that O->D = D->B.
O->B = a + c
so O->D = h(O->B) = h(a + c)
and similarly D->B = (1 - h) (a + c)
but
O->D + D->C = c and also A->D + D->B = c
which gives me
h(a+c) + (1-k)(c-a) = k(c-a) + (1-h) (a+c)
which is true but doesn't prove anything!
So I would be grateful for a hint. Thanks.
