For any right triangle ABC, the midpoint of the hypotenuse BC is equidistant from the 3 vertices A, B, C.

(Comment: This is one direction of the Carpenter Locus Theorem. The other direction says that if BC is a segment with midpoint O and if A is a point with OA = OB = OC, then angle BAC is a right angle.)

There are a number of correct answers, as we learned from our lab and write-up last Wednesday and also our discussion last Friday.

The typical error in an incorrect proof of this theorem is to implicitly assume the conclusion. For example, if we let O be the midpoint of BC and then say "construct OM the perpendicular bisector of AB," the write has assumed already thta O is on the perpendicular bisector of AB. But this is not yet known. In fact this is the key fact to be proved.

We are given a triangle ABC, with angle BAC a right angle. It is convenient to label the midpoint of the hypotenuse BC; let us call this midpoint O. From the definition of the midpoint, we know already that OB = OC.

If we can prove that OA = OB, we are done.

But we know that whenever we have a statement of equal distances such as OA = OB, geometrically this is the same thing as saying O is on the perpendicular bisector of AB (by the locus theorem for perpendicular bisectors). So if we can prove that O is on the perpendicular bisector of AB, we have OA = OB (as well as OB = OC), so we are done.

We will give a couple of ways to complete the proof.

Construct the line m which passes through O and which is perpendicular to AB. If we can prove that m is the perpendicular bisector of AB, we are done, since by construction, O is on m.

We know that m is perpendicular to AB. To show m is the perpendicular bisector, we need to show that it passes through the midpoint of AB. Let P be the intersection of m and AB. Then triangle BOP is similar to triangle BCA by AA, since the two triangles share one angle and they also both have angle BPO = angle BAC = right angle.

The ratio of similitude is BO/BC = ½, so by corresponding sides, it is also true that BP/BA = ½. This means that P is the midpoint of AB. Thus m, which is line OP, is the perpendicular bisector of AB.

Let M be the midpoint of AB. Construct line OM.

If we can prove that OM is the perpendicular bisector of AB, we are done, since O is on OM.

We know that OM passes through the midpoint M of AB. To show m is the perpendicular bisector, we need to show that it is perpendicular to AB. Then triangle BOM is similar to triangle BCA by SAS, since the two triangles share one angle and corresponding sides have the same ratio, BO/BC = BM/BA = ½, by the definition of midpoint.

So by corresponding angle, it is also true that angle BMO = angle BAC = right angle. This means that OM is perpendicular to AB. Thus OM is the perpendicular bisector of AB.

Let M be the midpoint of AB. Construct line n as the perpendicular bisector of AB. By definition, n passes through M. Let Q be the intersection of n and BC, so line n = line MQ.

We conjecture that Q is the midpoint O of BC. If this is true, then O is on the perpendicular bisector of AB and we are done. But we do not know (yet) that this is true, since Q was constructed as the intersection of two lines.

We know that triangle MBQ is similar to triangle ABC by AA, since they share an angle and angle CAB = angle QMB = right angle.

Then as before, the ratio of similitude is BM/BA = ½, so by corresponding sides, BQ/BC = ½ and Q is the midpoint of BC.

Thus Q = O and we are done, since we know OM is the perpendicular bisector of AB.