# Euclid Prop. 1

### On a given finite right line (AB) to construct an equilateral triangle.

### Sol.—With A as centre, and AB as radius, describe the circle BCD (Post. iii.).

With B as centre, and BA as radius, describe the circle ACE, cutting the former circle in C. Join CA, CB (Post. i.).

Then ABC is the equilateral triangle required.

Because A is the centre of the circle BCD, AC is equal to AB. Again, because B is the centre of the circle ACE, BC is equal to BA. Hence we have proved AC = AB, and BC = AB. Therefore AC is equal to BC; therefore the three lines AB, BC, CA are equal to one another. Hence the triangle ABC is equilateral; and it is described on the given line AB.

## Exercises.

The following exercises are to be solved when the pupil has mastered the First Book:--

### Rhombus

Casey's lozenge is what I call a rhombus. The following is from Wikipedia.

Traditionally, in two-dimensional geometry, a rhomboid is a parallelogram in which adjacent sides are of unequal lengths and angles are oblique.

A parallelogram with sides of equal length (equilateral) is a rhombus but not a rhomboid.

### 1. If the lines AF, BF be joined, the figure ACBF is a lozenge.

It can be shown that triangle ABF is equilateral by using above method.

As both triangles have common side AB, then all sides are equal, i.e.

AC = CB = BF = FA and ∠CAB = ∠FBA = 60∘ ∴ AF || BC (alt. ang.s equal) Draw CF and a similar argument will show AC || FB ∴ ABCD is a rhombus (lozenge) (opp. sides equal and parallel)

### 2. If AB be produced to D and E, the triangles CDF and CEF are equilateral.

∠ADC = ∠ACD (∆ADC isos. AD=DC) ∠BAC = ∠ADC +∠ACD (External ang. of tri. ADC) = 60∘ (proven above) ∴ ∠ADC = ∠ACD =30∘ ∠CAB = ∠FAB = 60∘ (∆ACB & ∆FAB both equilateral) ∴ ∠CAF = 120∘ ∠ACF + ∠AFC = 60∘ (angles of triangle = 180∘) ∠ACF = ∠AFC = 30∘ (isos. ∆ACF AC = AF) ∴ ∠DCF = ∠DFC = 60∘ ∴ ∠CDF = 60∘ (angles of triangle = 180∘) ∴ ∆CDF is equilateral (All 3 angles = 60∘) By similar argument it can be shown that ∆CFE is also equilateral

### 3. If CA, CB be produced to meet the circles again in G and H, the points G, F, H are collinear, and the triangle GCH is equilateral.

∠CAB = ∠BAC = 60∘ (proven above) ∴ ∠GAF = 60∘ (CAG is str. line = 180∘) ∠AGF = ∠CAB = 60∘ (corr. ang.s CB || AF) ∴ ∠AFG = 60∘ (angles of triangle = 180∘) ∴ ∆AGF is equilateral (all ang.s = 60∘) Similarly ∆BFG is equilateral ∴ ∠BHF = 60∘ ∠AFB = 60∘ (proven previously) ∴ GFH is str. line (∠GFA + ∠AFB + ∠BFH = 180∘) and ∆CGH is equilateral (all 3 ang.s = 60∘)

### 4. If CF be joined, CF² = 3AB².

CG² = GF² + CF² ∴CF² = CG² - GF² = CG² - AB² But CG² = 2AC² = 2AB² and CG² = 4AB² ∴ CF² = 4AB² - AB² = 3AB²

### 5. Describe a circle in the space ACB, bounded by the line AB and the two circles.

Draw EF parallel to BC Draw EI such that EI=EF and I lies on circle centred on A Join FI ∠EFI = ∠EIF (∆EFI isos.) Draw KE perpendicular to FI, K lying on FI ∠FEK = 180∘ - ∠EKF - ∠EFI = 180∘ - 90∘ - ∠EFI = 90∘ - ∠EFI = 90∘ - ∠EIF = ∠IEK ∴ ∆FKE ≣ ∆IKE (2 angles & included side =) ∴ FK = IK ∴ K lies on CD (angle bisector equi-distant from sides) ∠FKE = 90∘ = ∠BEK (Proven previously) ∴ FI || AB (Alternate angles =) ∠AEF = ∠EBF ( EF parallel to BC) = 60∘ ( Constructed previously) = ∠EFI ( alt. ang.s FI || AB) = ∠EIF ( proven) = ∠FEI (ang.s tri. = 180∘) ∴ ∆EFI is equilateral (all 3 ang.s = 60∘) Draw angle bisectors for ∠EFI & ∠EIF intercepting at point J J is centre of circle passing through E, F and I. (ang. bisectors of equil. tri. equi. dist from vertices) ∴ J is centre of required circle