Introduction to Euclid's Geometry — Practice

Question 1

Is 'the whole is greater than the part' an axiom, a postulate, or a theorem?

Accepted Answer

Step 1 — Recall the distinction.
  Axiom: a GENERAL statement accepted without proof, applies to anything.
  Postulate: a GEOMETRIC statement accepted without proof.
  Theorem: a statement derived (proved) from axioms/postulates.

Step 2 — Classify the given statement.
  'The whole is greater than the part' is a general statement about quantities, NOT specifically about geometry, and is accepted without proof.

✦ Answer: It is an axiom (specifically Axiom 5 in Euclid's list).

Question 2

Which postulate lets us extend a line segment indefinitely?

Accepted Answer

Step 1 — Read each postulate.
  P1: line through two points.
  P2: extend a line segment indefinitely.
  P3: circle.
  P4: right angles equal.
  P5: parallel postulate.

✦ Answer: Postulate 2 ('A terminated line can be produced indefinitely').

Why P2? Because a 'terminated line' = line segment, and 'produce' is old-mathematical English for 'extend'.

Question 3

Name three undefined terms in geometry.

Accepted Answer

Step 1 — Recall: some terms must be left undefined to avoid circular reasoning.

Step 2 — List the standard undefined terms.
  Point, line, plane.
  (We have intuitive notions of these but cannot define them without circular references.)

✦ Answer: Point, line and plane.

Why undefined? Try defining 'point' without using 'location', 'place', 'spot' — all of which themselves need definition. The chain has to stop somewhere.

Question 4

How many lines can pass through (a) one given point, (b) two distinct points?

Accepted Answer

Step 1 — Through one point.
  Imagine a point in space. You can swing a line through it in every possible direction.
  Number of lines: INFINITELY MANY.

Step 2 — Through two distinct points.
  Two points uniquely determine a line direction.
  By Postulate P1, EXACTLY ONE line passes through them.

✦ Answer: (a) Infinitely many. (b) Exactly one.

This is the geometric reason why we need TWO points (not one) to plot a unique line.

Question 5

True or false: 'Two distinct lines can intersect at more than one point.' Justify.

Accepted Answer

Step 1 — Suppose two distinct lines intersected at two points P and Q.

Step 2 — By Postulate P1, exactly ONE line passes through P and Q.

Step 3 — But we have TWO distinct lines both passing through P and Q — contradiction.

✦ Answer: FALSE. Two distinct lines can intersect at AT MOST one point. The proof uses Postulate P1.

Question 6

If AB = CD and CD = EF, prove that AB = EF.

Accepted Answer

Step 1 — Apply Axiom 1 (transitivity).
  Axiom 1: 'Things equal to the same thing are equal to one another.'
  Equivalently: if a = c and b = c, then a = b.

Step 2 — Set up the situation.
  Given: AB = CD and CD = EF.
  Both AB and EF are equal to the same thing (namely CD).

Step 3 — Conclude.
  By Axiom 1, AB = EF. ∎

✦ Answer: AB = EF, proved by Axiom 1.

Question 7

If AB = CD, prove that AB + BC = BC + CD.

Accepted Answer

Step 1 — Apply Axiom 2.
  Axiom 2: 'If equals are added to equals, the wholes are equal.'
  Adding the same thing (BC) to both sides preserves equality.

Step 2 — Apply to the given.
  AB = CD (given).
  Add BC to both sides: AB + BC = CD + BC.

Step 3 — Use commutativity of addition.
  CD + BC = BC + CD.

Step 4 — Combine.
  AB + BC = BC + CD. ∎

✦ Answer: AB + BC = BC + CD, proved using Axiom 2.

Question 8

If C is the midpoint of segment AB, prove that AC = (1/2) AB.

Accepted Answer

Step 1 — Definition of midpoint.
  C is the midpoint of AB ⇔ AC = CB AND C lies on segment AB.

Step 2 — Use the segment-addition postulate (or simply that AB = AC + CB if C lies between A and B).
  AB = AC + CB.

Step 3 — Substitute CB = AC (from Step 1).
  AB = AC + AC = 2 · AC.

Step 4 — Solve for AC.
  AC = AB / 2 = (1/2) AB. ∎

✦ Answer: AC = (1/2) AB, by definition of midpoint.

Question 9

State Euclid's Fifth Postulate and Playfair's Axiom. Show they are equivalent.

Accepted Answer

Step 1 — State Euclid's Fifth Postulate.
  If a straight line falling on two straight lines makes the interior angles on the same side together less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.

Step 2 — State Playfair's Axiom.
  Through a given point not on a given line, exactly one line can be drawn parallel to the given line.

Step 3 — Outline of equivalence.
  Given P5, we can derive Playfair (suppose two parallels exist through the same external point; use P5 to show they must coincide).
  Given Playfair, we can derive P5 (suppose the two lines do NOT meet; show this contradicts the unique-parallel statement).

✦ Answer: Both stated above. They are logically equivalent — assuming one allows derivation of the other within Euclidean axioms.

Question 10

Prove that two distinct lines cannot have more than one common point.

Accepted Answer

Step 1 — Proof by contradiction.
  Suppose, for contradiction, that two distinct lines ℓ and m have TWO common points, P and Q.

Step 2 — Apply Postulate P1.
  P1 says: through two distinct points exactly ONE line passes.
  But here we have ℓ and m both passing through P and Q.
  So ℓ and m must be the same line.

Step 3 — Reach contradiction.
  But ℓ ≠ m (given they are distinct). Contradiction.

Step 4 — Conclude.
  Our assumption (two distinct lines share two points) is false. ∎

✦ Answer: Two distinct lines have AT MOST one point in common.

Question 11

Prove: If a point C lies between A and B on a line, then AC + CB = AB. Use the axioms.

Accepted Answer

Step 1 — Set up.
  A, C, B are collinear and C is between A and B (given).

Step 2 — Use Axiom 4 (coincident things are equal).
  AC and CB together coincide with AB (when laid end to end starting at A and ending at B).
  By Axiom 4: AC + CB = AB.

Step 3 — Alternative argument using A2.
  If you add AC and CB to themselves (as line segments), you get the entire segment AB, by direct measurement.

✦ Answer: AC + CB = AB. This is sometimes called the 'segment-addition axiom' and is taken as a fundamental fact about collinear points.

Question 12

Why was Euclid's Fifth Postulate considered controversial? Briefly describe non-Euclidean geometry.

Accepted Answer

Step 1 — Why P5 stands out.
  P1–P4 are short, intuitive and obviously true. P5 is long and complicated.
  Mathematicians felt P5 should be PROVABLE from P1–P4 rather than assumed.
  For 2,000 years (300 BCE – 1820s CE), countless mathematicians tried — and failed.

Step 2 — The breakthrough (1820s).
  Gauss, Lobachevsky and Bolyai independently realised: P5 is INDEPENDENT of P1–P4.
  You can REPLACE P5 with a different axiom and get a different but logically consistent geometry.

Step 3 — Non-Euclidean geometries.
  Hyperbolic geometry: more than one parallel through an external point (saddle-shaped surfaces).
  Spherical / Elliptic geometry: no parallels — any two great circles on a sphere intersect (Earth's surface).

Step 4 — Modern significance.
  Einstein's General Relativity uses non-Euclidean geometry to describe gravity.
  GPS would be 11 km off per day if it used Euclidean approximations — modern technology critically depends on non-Euclidean geometry.

✦ Answer: P5 was independent of the other postulates. Replacing it gives non-Euclidean geometries used today in relativity and GPS.

Practice — Introduction to Euclid's Geometry

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities