Midpoint Theorum Help!!
I need help in this sum.
Q: In the adjoining figure, ABCD is a square and BCPQ is a parallelogram. If M is the midpoint of DC and PM produced meets AN at N, prove that AN=1/2 AD. If BD meets MN at R, prove that BQ=1/2 AC.
I have no problem with the first part of the question but the second part confuses me. Please help, thanks.
Please check the statement.
I think that it should be .
Originally Posted by Insaeno
I think it's more likely
Originally Posted by Plato
"If BD meets MN at Q, prove that BQ=1/2 AC."
It amounts to the same thing, but at least Q is on the diagram.
I'm pretty sure I copied the sum correctly. If it is so, then the question must be flawed. Thanks anyways!!!
This diagram also fits the given. Clearly in this diagram R could not be Q.
Letís suppose the R is the point of intersection of BD with NM (it may be Q or not). We know that NM is parallel to BC. In the triangle BCD, because M is the midpoint of CD then R is the midpoint of BD. The diagonals of a rectangle are the same length and bisect one another. Therefore, it follows that