Unit IV: Geometry — Test 1

(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.

(b) The intersection of its diagonals

1. The diagonals bisect each other at O.
2. A 180° turn about O exchanges both pairs of opposite vertices.

(c) 90°

1. AB subtends 180° at the centre.
2. The angle at C on the remaining circle is half this central angle, namely 90°.

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 union of AB, BC, CD, DA is a quadrilateral; opposite sides equal imply a parallelogram.

1. Join A to B to C to D to A, with four distinct coplanar vertices and no three collinear. A simple quadrilateral has no crossings other than adjacent sides meeting at vertices.
2. In simple ABCD draw AC. Given AB = CD and BC = DA, AC is common.
3. Triangles ABC and CDA are congruent by SSS.
4. Matching alternate interior angles imply AB ∥ CD and BC ∥ AD, giving a parallelogram.

PQ is parallel to BC and half its length.

1. Extend PQ through Q to S with QS = QP. Since AQ = QC and vertical angles at Q are equal, triangles AQP and CQS are congruent by SAS.
2. Hence AP = CS and AP ∥ CS. With AP = PB, the segment PB is equal and parallel to CS.
3. PBCS is a parallelogram, so PS ∥ BC and PS = BC.
4. P, Q, S are collinear and PS = 2PQ. Therefore PQ ∥ BC and PQ = BC/2.

Distance 3 cm; the perpendicular bisects AB.

1. Let M be the perpendicular foot. Right triangles OMA and OMB have equal hypotenuses OA = OB and common leg OM.
2. RHS gives AM = MB, so each half of the chord is 4 cm.
3. OM² = OA² − AM² = 25 − 16 = 9; OM = 3 cm.
4. Conversely, if M is the midpoint, SSS makes the adjacent angles at M equal and supplementary, hence each is 90°.

Intersect two perpendicular bisectors; the circumcentre is inside, at the hypotenuse midpoint, or outside respectively.

1. Construct perpendicular bisectors of AB and BC. Non-collinearity ensures they meet at a unique O.
2. O is equidistant from A and B, and from B and C, so OA = OB = OC.
3. Draw the circle of radius OA. Any centre of a circle through all three points must lie on both bisectors, proving uniqueness.
4. For an acute triangle O is inside; for a right triangle it is the hypotenuse midpoint; for an obtuse triangle it is outside.