A square sheet of paper ABCD is so folded that B falls on the midpoint M of CD.In which rtio the crease will divide BC. A square sheet of paper $ABCD$ is so folded that $B$ falls on the midpoint $M$ of $CD$. In which ratio will the crease divide $BC$?
 A: Let the crease be $QR$, with $Q$ on $BC$ and $R$ on $BM$. Then notice that $QR$ is the perpendicular bisector of $BM$. Then $\Delta BQR\cong\Delta MQR$.
WLOG, $QC=1,BQ=x$. Then $MC=\frac{1+x}2,MQ=x$. 
Then by Pythagoras on $\Delta MCQ$, we have $\left(\frac{1+x}2\right)^2+1=x^2$. Solving gives $x=\frac53$.
Thus, the crease divides $BC$ in a ratio of $5:3$.
$\boxed{\tiny Z}$
A: Let the side of the square be 2.

$\triangle BXY \sim \triangle BMC$. (Corrected. Originally was quoted as $\triangle BXY \sim \triangle AMC$.)
Setting up the corresponding ratios, we get $BX = \dfrac 54$ and hence $CX = 2 - BX = \dfrac 34$.

Extras (Completely re-written with the old picture replaced)

If, after folding, B falls on M, then the crease must be the perpendicular bisector of BM.
Let that perpendicular bisecor cut DC extended, BC, BM and AD at P, Q, R, S respectively.
By Pythagoras theorem, $MR = \dfrac {\sqrt 5}{2}$.
From $\triangle BQR \sim \triangle BMC$, we get $BQ = \dfrac {5}{4}$ and $QC = \dfrac {3}{4}$. In addition, QR is also known.
From $\triangle PCQ \sim \triangle PRM$, we get $PC = 1.5$.
From $\triangle PCQ \sim \triangle PDS$, we get $DS = \dfrac {7}{4}$.
Hence, $AS = 2 - DS = \dfrac 14$ and $DS : SA = 7 : 1$.
