### Fitting In

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

### Look Before You Leap

The diagonals of a square meet at O. The bisector of angle OAB meets BO and BC at N and P respectively. The length of NO is 24. How long is PC?

Two ladders are propped up against facing walls. The end of the first ladder is 10 metres above the foot of the first wall. The end of the second ladder is 5 metres above the foot of the second wall. At what height do the ladders cross?

# Two Triangles in a Square

##### Stage: 4 Challenge Level:

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.