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:—

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 = 60º

∴ 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 = ∠BAF = 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 verices)  ∴ J is centre of required circle