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 $BC$?

Let the side of the square be 2.

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$\triangle BXY \sim \triangle BMC$. (Corrected. Originally was quoted as $\triangle BXY \sim \triangle AMC$.)

Setting up the corresponding ratios, we get $BX = \dfrac 54$ and hence $CX = 2 - BX = \dfrac 34$.

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

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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 $DS : SA = 7 : 1$.