Statements of Euclid, Thales, similarity and the bisector theorem

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State, with a figure, hypothesis and thesis where appropriate:

• (a) Euclid’s first theorem, in equivalence and in similarity form, and its converse;
• (b) Euclid’s second theorem, in equivalence and in similarity form;
• (c) the three similarity criteria for triangles;
• (d) the small and the great Thales’ theorems, and the converse of the great one;
• (e) the angle bisector theorem.

Solution

Let $ABC$ be right-angled at $C$, $CH$ the altitude to the hypotenuse $AB$, with projections $AH,BH$ of the legs. (a) Euclid’s first theorem. Equivalence: the square on a leg equals the rectangle of the hypotenuse and that leg’s projection: $\overline{AC}^2=\overline{AB}\cdot\overline{AH}$. Similarity: each leg is the geometric mean between hypotenuse and its projection, $\overline{AB}:\overline{AC}=\overline{AC}:\overline{AH}$. Converse: if in a triangle a side is the geometric mean between a second side and the projection of the first onto the second, the triangle is right-angled. (b) Euclid’s second theorem. Equivalence: the square on the altitude to the hypotenuse equals the rectangle of the two projections: $\overline{CH}^2=\overline{AH}\cdot\overline{BH}$. Similarity: the altitude is the geometric mean between the two projections, $\overline{AH}:\overline{CH}=\overline{CH}:\overline{BH}$. (c) Similarity criteria. Two triangles are similar if: (1) they have two angles respectively congruent; (2) they have a congruent angle between proportional sides; (3) they have the three sides respectively proportional. (d) Thales. Small theorem: a line parallel to one side of a triangle divides the other two sides into proportional parts. Great theorem: a pencil of parallel lines determines proportional segments on two transversals. Converse of the great one: if a pencil determines proportional segments on two transversals (in order), then the lines are parallel. (e) Angle bisector theorem. The bisector of an interior angle divides the opposite side into parts proportional to the other two sides: $\dfrac{BD}{DC}=\dfrac{AB}{AC}$. $\boxed{\overline{AC}^2=\overline{AB}\cdot\overline{AH};\quad \overline{CH}^2=\overline{AH}\cdot\overline{BH};\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}.}$