# Midpoint Theorum Help!!

• Jul 14th 2007, 02:33 AM
Insaeno
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.
• Jul 14th 2007, 04:12 AM
Plato
I think that it should be $B{\color{red}R}=\frac{1}{2}AC$.
• Jul 14th 2007, 04:43 AM
topsquark
Quote:

Originally Posted by Insaeno
If BD meets MN at R, prove that BQ=1/2 AC.

Quote:

Originally Posted by Plato
I think that it should be $B{\color{red}R}=\frac{1}{2}AC$.

I think it's more likely
"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.

-Dan
• Jul 14th 2007, 05:35 AM
Insaeno
I'm pretty sure I copied the sum correctly. If it is so, then the question must be flawed. Thanks anyways!!!
• Jul 14th 2007, 07:50 AM
Plato
This diagram also fits the given. Clearly in this diagram R could not be Q.
Let’s 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 $BR=\frac{1}{2}AC$