23 Definitions

8. Angle

An angle is formed when two rays meet at a point.

Notation: ∠ABC.

9. Rectilinear Angle

An angle formed by two straight lines.

10. Right Angle

When a line meets another so that the two adjacent angles are equal, each is a right angle.

Notation: ∠ABC = 90° (π/2 rad).

11. Obtuse Angle

Greater than a right angle.

Notation: ∠ABC > 90°.

12. Acute Angle

Less than a right angle.

Notation: ∠ABC < 90°.

13. Boundary

The limit of a figure.

14. Figure

A shape enclosed by boundaries.

15. Circle

The set of all points at a fixed distance (radius) from a center O.

Notation: c(O,r) is the circle centered at O with radius r, that is, all points A such that |OA| = r.

16. Center of Circle

The point O from which all radii are equal.

17. Diameter

A line through the center, ending on the circle.

Notation: AB with O midpoint; length = 2r.

18. Semicircle

Half a circle, bounded by a diameter and the arc it cuts off.

Notation: arc(AB).

19. Rectilinear Figure

A figure bounded by straight lines (a polygon).

20. Triangle

A polygon with three sides.

Notation: △ABC.

21. Quadrilateral

A polygon with four sides.

Notation: □ABCD.

22. Polygon

A rectilinear figure with more than four sides.

23. Types of Triangles

By sides: equilateral (three equal), isosceles (two equal), scalene (none equal).
By angles: right (one right), obtuse (one obtuse), acute (all acute).

Notation: equilateral |AB| = |BC| = |CA|; isosceles |AB| = |AC|; right ∠ABC = 90°, etc.

5 Postulates

1. Line between Points

A straight line can be drawn joining any two points.

Notation: segment AB.

2. Extension of Line

A finite line can be extended indefinitely.

Notation: extend AB to C.

3. Circle

A circle can be drawn with any center and radius.

Notation: c(O,r).

4. Right Angles

All right angles are equal.

Notation: if ∠ABC = 90° and ∠DEF = 90°, then they are equal.

5. Parallel Postulate

Given a line and a point not on it, exactly one line can be drawn through the point parallel to the given line.

Notation: through P not on ℓ, there is a unique m with m ∥ ℓ.

5 Common Notions

1. Equality (Transitive)

Things equal to the same thing are equal to one another.

Notation: if a = b and b = c, then a = c.

2. Addition Rule

If equals are added to equals, the results are equal.

Notation: if a = b, then a + c = b + c.

3. Subtraction Rule

If equals are subtracted from equals, the remainders are equal.

Notation: if a = b, then a − c = b − c.

4. Coincidence Principle

Things which coincide with one another are equal to one another.

Notation: if △ABC ≅ △DEF, then all corresponding parts equal.

5. Whole vs. Part

The whole is greater than the part.

Notation: if B lies between A and C, then |AB| < |AC|.

Equilateral Triangle

An equilateral triangle can be constructed on any given finite straight line.

ElementsI.1 Take a stick and mark its ends A and B. With your compass on A, swing a circle through B. Then with your compass on B, swing a circle through A. Where the circles meet, call that point C. Join A to C and B to C. You’ve made a triangle with all three sides equal.

Proof

By the definition of a circle, every point on c_A is at distance |AB| from A; in particular, |AC| = |AB|. Likewise, every point on c_B is at distance |AB| from B; in particular, |BC| = |AB|. Hence |AC| = |BC| = |AB|, so the sides of △ABC are equal and the triangle is equilateral. Q.E.D.

Construction

1. Let AB be the given segment.
2. Draw circle c_A centered at A with radius |AB|.
3. Draw circle c_B centered at B with radius |AB|.
4. Let C be a point of intersection: C = cA ∩ cB, with C ≠ A,B.
5. Join AC and BC to form △ABC.

Triangle Congruences

Side–Angle–Side (SAS)

If two triangles have two pairs of corresponding sides equal and the included angles equal, then the triangles are congruent.

ElementsI.4 Make a triangle by hinging two sticks at a fixed angle. If a second triangle uses sticks of the same lengths and the same hinge angle, the two triangles match exactly when you lay one on top of the other.

Proof

Let △ABC and △DEF satisfy |AB| = |DE|, |AC| = |DF|, and ∠BAC = ∠EDF. Superpose △ABC onto △DEF by placing A onto D and aligning AB with DE. Because the included angles at A and D are equal, ray AC coincides with ray DF, and since |AC| = |DF|, point C falls on F. Thus △ABC and △DEF coincide: △ABC ≅ △DEF. Corresponding bases and angles are equal. Q.E.D.

Side–Side–Side (SSS)

If two triangles have all three pairs of corresponding sides equal, then the triangles are congruent.

ElementsI.8 Build two triangles from the same three stick lengths. There’s only one way (up to flipping) to join them: the triangles match.

Proof

Let △ABC and △DEF satisfy |AB| = |DE|, |BC| = |EF|, |CA| = |FD|. Place A on D with AB along DE. The locus of points at distance |CA| from A is a circle; likewise the locus at distance |FD| from D is the same radius circle. The equalities force the third vertex to the same intersection point on the same side of the base, so C falls on F. Hence △ABC ≅ △DEF. Corresponding angles are equal. Q.E.D.

Angle–Side–Angle (ASA) and AAS

If two triangles have two angles equal respectively and one corresponding side equal (either the included side or a side opposite one of the equal angles), then the triangles are congruent.

ElementsI.26 Hinge two angles the same way and make one side the same; the third vertex lands in only one place, so the triangles match.

Proof

Let △ABC and △DEF satisfy ∠A = ∠D, ∠B = ∠E and either |AB| = |DE| (adjacent case) or |AC| = |DF| (opposite case). Equal two angles force the third angles equal. In the adjacent case, the included angle between the given sides is equal; Side-Angle-Side applies. In the opposite case, placing the equal side and opening the equal adjacent angles determines the remaining side uniquely, yielding superposition of the triangles. Thus △ABC ≅ △DEF. Q.E.D.

Isosceles Triangle

Base Angles

In any triangle with two equal sides, the angles at the base are equal.

ElementsI.5 Make a triangle where the two sides from the top point are the same length—like two equal sticks hinged at the top. When you set the base on the table, the “feet” sit symmetrically. The tilt at the left foot matches the tilt at the right foot.

Proof

Let △ABC satisfy |AB| = |AC|. Compare △ABC with △ACB (swap the labels of B and C). Then the side pairings are |AB| ↔ |AC| and |AC| ↔ |AB| (equal by the given), and the included angle is ∠BAC in both triangles. By SAS congruence, △ABC ≅ △ACB. Therefore the base angles correspond and are equal: ∠ABC = ∠ACB. Q.E.D.

Converse: Opposite Sides

In any triangle, if two base angles are equal, then the opposite sides are equal.

ElementsI.6If a triangle “leans” the same amount at both feet, then the two side sticks up to the top must be the same length—otherwise one side would reach higher or lower and spoil the symmetry.

Proof

Let △ABC satisfy ∠ABC = ∠ACB. Compare △ABC with △ACB (swap B and C). They have two equal angles (∠A is common; ∠B = ∠C by the given) and the corresponding side |BC| = |CB|. By ASA/AAS congruence, △ABC ≅ △ACB. Therefore the sides opposite the equal base angles are equal: |AB| = |AC|. Q.E.D.

Bisection

Segment Bisection

Any finite straight line can be cut into two equal parts.

ElementsI.10 Take a stick AB. Swing equal-radius arcs from A and B that cross above and below the stick. The line through the crossings meets AB at its midpoint M; it’s also ⟂ to AB.

Proof

Let circles with the same radius meet at X and Y. Then |XA| = |XB| and |YA| = |YB|. In △XAB and △YAB, equal radii give two equal sides; the base angles at A and B match, so XY is the perpendicular bisector of AB. Hence the intersection M = XY ∩ AB satisfies |AM| = |MB|. Q.E.D.

Construction

1. Let AB be given.
2. Draw circle c_A centered at A with any radius > |AB|/2.
3. Draw circle c_B centered at B with the same radius.
4. Let X, Y be the two intersections: {X,Y} = cA ∩ cB.
5. Draw the line XY; let M = XY ∩ AB. Then |AM| = |MB| and XY ⟂ AB.

Angle Bisection

Any rectilinear angle can be cut into two equal angles.

ElementsI.9 From the angle’s vertex A, swing a circle to mark points E on AB and F on AC. With the same opening, swing arcs from E and F to meet at D. The ray AD bisects the angle.

Proof

With |AE| = |AF| and equal radii from E and F, triangles △AED and △AFD have two sides equal and the included side AD common, so by SAS they are congruent. Hence ∠EAD = ∠DAF. Q.E.D.

Construction

1. Given rays AB and AC with common endpoint A.
2. Draw circle c_A centered at A to meet the rays at E (on AB) and F (on AC).
3. With the same radius, draw circles c_E and c_F.
4. Let D be an intersection of c_E and c_F inside the angle.
5. Draw AD; then ∠EAD = ∠DAC.

Perpendiculars

Perpendicular at a Point on a Line

From a point on a given straight line, a straight line can be drawn at right angles to the given line.

ElementsI.11 On the line ℓ, pick equal points B and C on either side of A. With the same radius, swing arcs from B and C to meet at X and Y. The line XY passes through A and is ⟂ to ℓ.

Proof

From the equal-radius circles at B and C, we have |XB|=|XC| and |YB|=|YC|. Thus X and Y lie on the perpendicular bisector of BC. Since A is the midpoint of BC, the line through X and Y is ⟂ to ℓ at A. Q.E.D.

Perpendicular from a Point to a Line

From a point not on a given straight line, a straight line can be drawn perpendicular to the given line.

ElementsI.12 With center P, swing a circle to meet the line at X and Y. Bisect XY at M. The line PM is ⟂ to the base line.

Proof

The points X and Y are equidistant from P. The midpoint M of XY lies on the perpendicular bisector of XY. Since P is also equidistant from X and Y, the line PM is that perpendicular bisector, hence ⟂ to the base line at M. Q.E.D.

Exterior Angle Theorem

In any triangle, if a side is produced, the exterior angle is greater than either of the two non-adjacent interior angles.

ElementsI.16 For △ABC, extend side BC to D. Then the exterior angle at C satisfies ∠ACD > ∠A and ∠ACD > ∠B.

Proof

Bisect AC at E (1.4). Join BE, and produce it beyond E to F so that BE = EF. Join FC. Then AE = CE, BE = EF, and the vertical angles at E are equal (I.15), so △ABE ≅ △CEF (SAS, I.4). Hence ∠A = ∠ECF. Since ray CF lies inside the exterior angle ∠ACD, we have ∠ECF < ∠ACD, so ∠A < ∠ACD. Repeating the same construction with the midpoint of BC shows ∠B < ∠ACD. Q.E.D.