Prepliminary note:
Let PQRS be a parallelogram, with T a point on one of its diagonals, say PR. Lines are constructed through T parallel to PQ and PS. Now the two smaller parallelograms with vertices at Q and S, and sharing a common vertex at T, are obviously of the same area.
Proof.
Turning to the given problem. if and only if
D is the midpoint of side BC, will lines through D parallel to AB and AC pass through the midpoints of AC and AB respectively. A figure of the sort mentioned in the preliminary note is now completed by constructing a second set of parallels through E and F, with T again a point at the crossing between this latter pair of parallels.
We have already seen from the preliminary note that the parts of BDF and CED that lie inside the parallelogram are equal, being bisections of the smaller parallelograms; and the remaining portions of those triangles are clearly congruent. It is equally obvious that this equality applies if and only if D is at the midpoint, for otherwise T will no longer lie on AD, and the size of one triangle will increase or decrease at the expense of the other.
I'm not sure what is meant by 'orientation'; for example, if T is rather close to A, with a right angle at A, and the length of AB is, say, 10AC, the triangles can be made to look quite different.
Edited on February 20, 2012, 1:28 am
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Posted by broll
on 2012-02-19 08:02:30 |
Possible solution
