1. ## 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.

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

3. Originally Posted by Insaeno
If BD meets MN at R, prove that BQ=1/2 AC.
Originally Posted by Plato
I think that it should be $\displaystyle B{\color{red}R}=\frac{1}{2}AC$.
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

4. I'm pretty sure I copied the sum correctly. If it is so, then the question must be flawed. Thanks anyways!!!

5. 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 $\displaystyle BR=\frac{1}{2}AC$