Pollock2

Pollock's theorem construction

"c" the circumcircle of the orthogonal triangle ABC.

i) If point Z is such that the tangent segment E'D' is bisected at Z, then an easy calculation shows that v = <(ZCD') must be of measure (pi-a)/3, where a = <(CAB).
Thus to construct point Z:
i1) Select [Measures\Angle Label] and click on angle A. This produces a label of the form [ <(A) XX:XX], indicating the measure a of angle A.
i2) Write somewhere the text [formula (pi-x)/3] and press the enter key. This produces the formula object [(pi-x)/3].
i3) Right-click on the formula-object and select [Activate]. Then click on the label of a1) to produce the measure-label of the angle u = (pi-a)/3. This produces a box of the form [Eval YY:YY].
i4) Double click on the last created label-box, while holding down the [shift] key. This produces the angle v, whose measure is (pi-a)/3.
i5) Select the tool [Angle Compasses _ _ _ ] and click on v and then on the anchors C and B to transfer the angle v at C, along CB. This creates a segment on the direction of CZ.
i6) Select the line-tool and create the line CZ. Find its intersection Z with the circle.

ii) If point I is such that the tangent segment KL is bisected a I, then an easy calculation shows that u = <(ICB) must be of measure a/3, where a = <(CAB). Construct I by the method of a), replacing the formula [(pi-x)/3] there, with the formula [x/3].

iii) The same construction with (ii) can be caried out for the point M, such that the tangent segment PQ is bisected at M.

Pollock's theorem says that ZIM is an equilateral triangle. The proof is an easy consequence of the previous angle relations.

http://users.math.uoc.gr/~pamfilos/