A square sheet of paper ABCD is so folded that B falls on the mid-point of M of CD. Prove that the crease will divide BC in the ratio 5:3.

Discussion:

Assuming the side of the square is ‘s’. Let a part of the crease be ‘x’ (hence the remaining part is ‘s-x’). We apply Pythagoras Theorem we solve for x:

\(x^2 + \frac {s^2 }{4} = (s-x)^2 \) implies \(x = \frac {3s}{8} \) and \(s-x = \frac {5s}{8} \)

Hence the ratio is 5:3.

## One Reply to “Crease of a square paper”