Geometry Foundations: Euclid, Lines and Triangles Worksheet

(c) Reflex

1. A reflex angle is greater than 180° and less than 360°.
2. 230° satisfies these inequalities.

(d) SSA

1. SSS, SAS and ASA determine a triangle up to rigid motion.
2. Two sides and a non-included angle can describe two different triangles; SSA needs additional conditions.

(c) 142°

1. A linear pair sums to 180°.
2. Subtract 38° to obtain 142°.

Every non-degenerate triangle has at least two acute angles.

1. The three positive interior angles sum to 180°.
2. Two angles each at least 90° would leave at most 0° for the third. In particular two obtuse angles are impossible.
3. If all three were less than 60°, their sum would be less than 180°, another contradiction.

SAS; A ↔ P, B ↔ Q, C ↔ R; BC = QR.

1. The equal angle is included between the two specified equal sides in each triangle.
2. SAS gives ABC congruent to PQR with the stated correspondence.
3. Corresponding remaining sides therefore give BC = QR. An approximate drawing is not the justification.

Practical measurement and construction motivated geometry; definitions specify meanings and axioms/postulates are accepted starting rules.

1. Land measurement, building and ritual constructions supplied practical geometric problems; Greek deductive geometry organised results into proofs.
2. Definitions state meanings, for example a circle as points at fixed distance from a centre. Axioms are general accepted principles; postulates concern permitted geometric objects and constructions.
3. Postulates 1 and 2: draw a straight segment joining any two points; extend a finite straight segment indefinitely.
4. Postulates 3 and 4: draw a circle with any centre and positive radius; all right angles are equal.
5. Postulate 5: if a transversal makes same-side interior angles sum to less than two right angles, the two lines meet on that side when extended.

Construct AD and BC perpendicular to AB on the same side, with AD = BC = AB; join C to D.

1. Construct perpendicular rays at A and B on the same side of AB.
2. Transfer length AB onto these rays to mark D and C.
3. AD and BC are equal and parallel, so ABCD is a parallelogram. A right angle makes it a rectangle.
4. Adjacent sides AB and AD are equal, hence it is a square. This illustrates the square-construction problem found in the Sulbasutra tradition; it is not a claim that this is the sole historical procedure.

A ray has a starting point and extends in one direction; an angle is formed by two rays with a common endpoint. Vertically opposite angles are equal.

1. A linear pair consists of adjacent angles whose non-common arms are opposite rays; their sum is a straight angle, 180° = two right angles.
2. Conversely, if two adjacent angles sum to 180°, their non-common arms form a straight line.
3. For intersecting lines, let consecutive angles be α, β, γ. Linear pairs give α + β = 180° and β + γ = 180°.
4. Subtract equal β from the equal sums to obtain α = γ. Repeat for the other pair.

65°, 65°, 115° respectively.

1. Corresponding and alternate interior angles are equal for parallel lines, so both are 65°.
2. Same-side interior angles are supplementary: 180° − 65° = 115°.
3. Conversely, equality of corresponding or alternate interior angles, or supplementary same-side interior angles, implies parallelism.
4. Distinct lines parallel to the same line in a plane are parallel to each other. Parallel lines are coplanar and do not meet.

A triangle’s three side lengths fix its shape; ∠ABC = ∠BCA.

1. Three side lengths determine a triangle by SSS, whereas four side lengths alone do not fix a quadrilateral’s angles. Diagonal braces split a frame into rigid triangles.
2. Compare triangles ABC and ACB: AB = AC, AC = AB and BC = CB.
3. SSS gives congruence with A ↔ A, B ↔ C and C ↔ B.
4. Corresponding base angles at B and C are equal. This is why triangular bracing is useful in structures.

RHS applies specifically to right triangles with equal hypotenuse and one corresponding leg.

1. SAS: two corresponding sides and the included angle; SSS: three corresponding sides.
2. ASA: two corresponding angles and the included side. AAS: two corresponding angles and any corresponding non-included side; the third angles are equal.
3. RHS: two right triangles have equal hypotenuses and one corresponding leg. The other leg follows from Pythagoras, reducing the problem to SSS.
4. General SSA lacks a right-angle/hypotenuse condition and may be ambiguous. The correspondence and included-angle requirement must be checked.

The converse is false; a non-square rectangle is a counterexample. RHS is a valid special situation.

1. The converse is: if all angles of a quadrilateral are equal, it is a square.
2. A rectangle with sides 2 cm and 3 cm has four equal angles but is not a square.
3. For SSA, a circle fixing one side can meet the ray fixing another angle at two points, producing two possible triangles. An approximate drawing cannot settle congruence.
4. For right triangles with equal hypotenuse and one equal leg, RHS fixes the remaining leg and gives congruence.

The two lines are parallel to each other.

1. Let l and m be distinct and each parallel to n.
2. Assume l and m intersect at P.
3. Both would be lines through P parallel to n, contradicting the uniqueness of a parallel through a point outside n.
4. Therefore they do not intersect. Coplanarity and distinctness are essential to the statement.