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Proposition - 1
Raman Choubay
Proposition 1โ€ข To construct an equilateral triangle on a given finite
straight-line.
โ€ข Step for construction
Describe the circle BCD with centre A and radius AB.
Again describe the circle ACE with centre B and radius BA.
Join the straight lines CA and CB from the point C at which the circles cut one another to
the points A and B
Proof
Statements Reason
AB = AC โˆต ๐ด ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘๐‘–๐‘Ÿ๐‘๐‘™๐‘’๐ถ๐ท๐ต
BA = BC โˆต ๐ต ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘๐‘–๐‘Ÿ๐‘๐‘™๐‘’ ๐ถ๐ด๐ธ
โ‡’ ๐ด๐ถ = ๐ถ๐ต โˆต ๐ด๐ต = ๐ด๐ถ ๐‘Ž๐‘›๐‘‘ ๐‘Ž๐‘™๐‘ ๐‘œ ๐ด๐ต = ๐ต๐ถ ๐‘Ž๐‘›๐‘‘
Things equal to the same thing are
also equal to one another
AB = AC = BC โˆต ๐ด๐ต = ๐ด๐ถ, ๐ต๐ด = ๐ต๐ถ ๐ด๐‘๐ท ๐ด๐ถ = ๐ถ๐ต
Thus, the triangle ABC is equilateral, and has been constructed on the given finite
straight-line AB.
Q.E.F

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Book 1 prop-1_equilateral triangle

  • 2. Proposition 1โ€ข To construct an equilateral triangle on a given finite straight-line. โ€ข Step for construction Describe the circle BCD with centre A and radius AB. Again describe the circle ACE with centre B and radius BA. Join the straight lines CA and CB from the point C at which the circles cut one another to the points A and B
  • 3. Proof Statements Reason AB = AC โˆต ๐ด ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘๐‘–๐‘Ÿ๐‘๐‘™๐‘’๐ถ๐ท๐ต BA = BC โˆต ๐ต ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘๐‘–๐‘Ÿ๐‘๐‘™๐‘’ ๐ถ๐ด๐ธ โ‡’ ๐ด๐ถ = ๐ถ๐ต โˆต ๐ด๐ต = ๐ด๐ถ ๐‘Ž๐‘›๐‘‘ ๐‘Ž๐‘™๐‘ ๐‘œ ๐ด๐ต = ๐ต๐ถ ๐‘Ž๐‘›๐‘‘ Things equal to the same thing are also equal to one another AB = AC = BC โˆต ๐ด๐ต = ๐ด๐ถ, ๐ต๐ด = ๐ต๐ถ ๐ด๐‘๐ท ๐ด๐ถ = ๐ถ๐ต Thus, the triangle ABC is equilateral, and has been constructed on the given finite straight-line AB. Q.E.F