The carpenter's problem

A carpenter has to join two thin strips of wood in a cross, so that a thread passed round their four ends makes a rectangle. She has one strip 8 cm long. How long should the other be, and where should they be joined?

First, model it. The strips are line segments, and they are the diagonals of the quadrilateral whose corners are their ends. So three things decide everything:

• how long is the other diagonal?
• where do the two cross?
• at what angle?

First, a tool. Two triangles are congruent (≅) when one fits exactly on top of the other, like your two hands: same shape and same size. You can tell they're congruent without measuring everything, using conditions such as SAS (two sides and the angle between them), AAS (two angles and a side) or SSS (all three sides). Once they're congruent, every matching part is equal, which is how all the proofs in this chapter work.

Deduction 1: the length of the other diagonal. Suppose ABCD is already a rectangle, and compare ∆ADC and ∆DAB:

• AB = CD, because opposite sides of a rectangle are equal;
• ∠BAD = ∠CDA = 90°, because all its angles are right angles;
• AD is in both triangles.

That's two sides and the angle between them, so by SAS, ∆ADC ≅ ∆DAB. So AC = BD, as matching parts of congruent triangles. The diagonals of a rectangle are always equal, so the second strip must also be 8 cm.

Deduction 2: where they cross. Let the diagonals meet at O, and compare ∆AOB and ∆COD. The angles at O are equal, being vertically opposite. AB = CD as before. What about ∠1 and ∠2?

Follow the angles round. Since ∠B = 90°, ∠3 + ∠1 = 90°. And in ∆BCD, ∠3 + ∠2 + 90° = 180°, so ∠3 + ∠2 = 90°. Both ∠1 and ∠2 equal 90° − ∠3, so ∠1 = ∠2.

Now by AAS, ∆AOB ≅ ∆COD. So OA = OC and OB = OD: O is the midpoint of both diagonals. When diagonals cross at their midpoints we say they bisect each other. To bisect means to cut into two equal parts (bi = two, like the two wheels of a bicycle). So the carpenter must join the strips at their middles.