# Folded Square

This square piece of paper has been folded and creased. Where does the crease meet the side AD?

## Problem

This 8 cm by 8 cm square of paper has been folded so that the corner at B lies exactly on top of E, the midpoint of CD. The dotted line shows the crease left by this fold.

Image

How far from A does the dotted line reach the side AB?

## Student Solutions

In the diagram below, the line EB has been added. The dotted line must be the perpendicular bisector of EB, since for B to fold onto E, B and E must be the same distance on either side of the fold.

The distance CE is half of the distance BC, so the gradient of EB must be 2.

This means that the gradient of the dotted line must be $-\frac{1}{2}$.

”‹The dotted line and the folded line cross halfway between E and B (since EB is bisected), so 4 cm from the top (and from the bottom) and 2 cm from the right (and from E). So the distance marked on the diagram below is 6 cm.

So the vertical distance between the point where the triangle meets AD and the point where the dotted line and the folded line cross is $\frac{1}{2}\times$6 = 3 cm.

So the dotted line meets AD 4$-$3 = 1 cm below A.

The two triangles coloured blue are similar, as they both contain a right angle and their other two angles are 90$^\text{o}$ rotations of each other.

The dotted line and the folded line cross halfway between E and B (since EB is bisected), so 4 cm from the top (and from the bottom) and 2 cm from the right (and from E). So the distance marked on the diagram below is 6 cm.

The sides of the larger triangle are in the ratio 1:2 (from 4:8), so the vertical side of the smaller triangle must be 3 cm. So the dotted line meets AD 4$-$3 = 1 cm below A.

Image

**Using gradient**The distance CE is half of the distance BC, so the gradient of EB must be 2.

This means that the gradient of the dotted line must be $-\frac{1}{2}$.

”‹The dotted line and the folded line cross halfway between E and B (since EB is bisected), so 4 cm from the top (and from the bottom) and 2 cm from the right (and from E). So the distance marked on the diagram below is 6 cm.

Image

So the vertical distance between the point where the triangle meets AD and the point where the dotted line and the folded line cross is $\frac{1}{2}\times$6 = 3 cm.

So the dotted line meets AD 4$-$3 = 1 cm below A.

**Using similar triangles**The two triangles coloured blue are similar, as they both contain a right angle and their other two angles are 90$^\text{o}$ rotations of each other.

The dotted line and the folded line cross halfway between E and B (since EB is bisected), so 4 cm from the top (and from the bottom) and 2 cm from the right (and from E). So the distance marked on the diagram below is 6 cm.

Image

The sides of the larger triangle are in the ratio 1:2 (from 4:8), so the vertical side of the smaller triangle must be 3 cm. So the dotted line meets AD 4$-$3 = 1 cm below A.