Congratulations to all those who submitted
solutions. Billy from Sedgehill School drew a diagram showing sides
and angles he knew must be equal, which helped him to spot 30 and
60 degree angles. Patrick from Woodbridge School went a little
further and made a good argument using ideas about congruent
triangles. Jack's (Fast Forward) solution was also very clear and
used similar ideas to Patrick's; I have combined their solutions
I know that KEC is a right angle because it is the corner of the
paper, so as KH is a straight line, CEH is also a right angle. The
line KE = EH; I know this because E lies on FG which divides the
sheet in two. EC is common to both triangles, so triangles KEC and
CEH are congruent (Side-Angle-Side).
I know that triangle KDC is also congruent to KEC because we folded
corner D up to E to create the two triangles.
As the three triangles are congruent, I know that angles DCK, KCE
and ECH are all equal, so as DCH is a right angle, each angle must
be $90 \div 3$ which is $30$ degrees.
Angles in a triangle add up to $180$ degrees, so angle CKE and CHE
are $180-90-30=60$ degrees, so triangle CKH is equilateral.