Or search by topic
You may like to ponder the fact that, looking at David's diagram below, the square LPKJ is 1/5th of the area of the square ABCD.
The following solution came from David (Madras College).
Label angles x and y, where x + y = 90^{\circ}, \angle MBC = x and \angle ABM = y
A line is drawn from point A to point L so that \angle ALB = 90^{\circ}
Triangles ABL and BCP are congruent because they have equal angles and AB = BC. Similarly triangles CDK and DAJ can be proven congruent with ABL and BCP forming a "camera shutter'' shape.
Triangle ABM is similar to the four congruent triangles (with angles x, y and 90^{\circ}) and therefore the proportions are the same.
Because AB = 2AM we have PC = 2PB and then, as KC = PB, we have PC = 2KC, so K is the midpoint of the line PC.
When triangle CDK is reflected along the line DK, the line DP is the reflection of the line DC and they are therefore the same length.
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ
You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle?