# Pythagoras' from Bottema's

Bui Quang Tuan found a way to derive the Pythagorean theorem from an elegant lemma of his. The lemma can be also used to deduce the Pythagorean theorem from that of Bottema. The applet below illustrates the latter derivation.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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Let ΔABC be right at C. Erect right isosceles triangles AA'C and BB'C externally to ΔABC. According to Bottema's theorem, the midpoint M of A'B' is the apex of the isosceles right triangle ABM (and thus is independent of C.) Point C also lies on A'B'. (This is because the angles at C add up to 180°: 45° + 90° + 45° = 180°.)

The configuration is exactly that of Bui Quang Tuan's lemma: AA'||BC and BB'||AC. Which allows one to conclude that:

Area(AA'C) + Area(BB'C) = 2 Area(AMB).

Letting, as usual, a = BC = B'C, b = AC = A'C, and c = AB = 2·OM, we can express the equality of the areas as

b²/2 + a²/2 = 2·c/2·c/2,

which is the Pythagorean identity.

The same configuration admits a different interpretation.

For a right ΔABC, the circumcircle (O) is centered at the midpoint O of the hypotenuse AB and has the radius of c/2. Let M be the midpoint of arc ACB. Form circles A(C) centered at A and B(C) centered at B, both passing through C. Note points A' and B' of intersection of line CM with the two circles.

In circle (O), arc(AM) = arc(BM) = 90°; hence ∠ACA' = ∠BCB' = 45°. This means that triangles AA'C and BB'C are right isosceles at A and B, respectively. We are now in position to apply a theorem of two intersecting circles and then, as before, the area lemma. ### Bottema's Theorem 