1. Proposition 33
Raman choubay

2. Proposition 33
To draw a segment of a circle, accepting an angle equal to a given rectilinear angle, on a given straight-line.

3. So the angle C is surely either acute, a right-angle, or obtuse.
Proposition 33
Given: Let AB be the given straight-line, and C the given rectilinear angle.
So it is required to draw a segment of a circle, accepting an angle equal to C, on the given straight-line AB.
So the angle C is surely either acute, a right-angle, or obtuse.

4. Case-1 Acute A B
let ∠ BAD= ∠ C, have been constructed at the point A on the straight line AB D A F
Let AE have been drawn, at right-angles to DA G B E
let AB have been cut in half at F
let F G have been drawn from point F , at right-angles to AB
G is the centre of the segment and GB is radius Join GB and draw the circle

5. Proof case-1 Statements Reason 𝐼𝑛 ∆ 𝐴𝐺𝐵 ∠𝐺𝐹𝐴=∠𝐺𝐹𝐵 𝐵𝑦 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
𝐴𝐹=𝐹𝐵
𝐵𝑦 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
𝐹𝐺+𝐹𝐺
𝐶𝑜𝑚𝑚𝑜𝑛
∆𝐴𝐺𝐹 ≅∆𝐵𝐺𝐹
By SAS
𝐵𝐺=𝐴𝐺
𝐶𝑃𝐶𝑇
𝐴𝐷 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑡 𝐴
𝐴𝐷 ⊥𝐴𝐸 𝑎𝑛𝑑 𝑐𝑒𝑛𝑡𝑟𝑒 𝐺 𝑖𝑠 𝑜𝑛 𝐴𝐸
𝐴𝑑 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑛𝑑 𝐴𝐵 𝑖𝑠 𝐶ℎ𝑜𝑟𝑑 𝑠𝑜 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑠𝑖𝑑𝑒 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙
∠𝐷𝐴𝐵=∠𝐴𝐸𝐵
∠𝐶=∠𝐴𝐸𝐵
𝐵𝑦 𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
Thus, a segment AEB of a circle, accepting the angle AEB (which is) equal to the given (angle) C, has been drawn on the given straight-line AB..
QED

6. Case-2 Right A B
let ∠ BAD= ∠ C, have been constructed at the point A on the straight line AB D A F
let AB have been cut in half at F B
let the circle AEB have been drawn with center F , and radius either F A or F B

7. Proof case-2 Statements Reason 𝐴𝐷 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑡 𝐴
𝐴𝐷 ⊥𝐴𝐵 𝑎𝑛𝑑 𝑐𝑒𝑛𝑡𝑟𝑒 𝐺𝐹 𝑖𝑠 𝑜𝑛 𝐴𝐵
∠𝐷𝐴𝐵= 90 0
𝐴𝐷 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡
∠𝐴𝐸𝐵= 90 0
𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒
∠𝐴𝐸𝐵=∠𝐷𝐴𝐵=∠𝐶
𝐵𝑦 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑎𝑏𝑜𝑣𝑒
Thus, a segment AEB of a circle, accepting an angle equal to C, has been drawn on AB
QED

8. Case-3 obtuse A B D
let ∠ BAD= ∠ C, have been constructed at the point A on the straight line AB A F
Let AE have been drawn, at right-angles to AD G B E
let AB have been cut in half at F
let F G have been drawn from point F , at right-angles to AB
G is the centre of the segment and GB is radius Join GB and draw the circle

9. Proof case-1 Statements Reason 𝐼𝑛 ∆ 𝐴𝐺𝐵 ∠𝐺𝐹𝐴=∠𝐺𝐹𝐵 𝐵𝑦 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
𝐴𝐹=𝐹𝐵
𝐵𝑦 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
𝐹𝐺+𝐹𝐺
𝐶𝑜𝑚𝑚𝑜𝑛
∆𝐴𝐺𝐹 ≅∆𝐵𝐺𝐹
By SAS
𝐵𝐺=𝐴𝐺
𝐶𝑃𝐶𝑇
𝐴𝐷 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑡 𝐴
𝐴𝐷 ⊥𝐴𝐸 𝑎𝑛𝑑 𝑐𝑒𝑛𝑡𝑟𝑒 𝐺 𝑖𝑠 𝑜𝑛 𝐴𝐸
𝐴𝐷 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑛𝑑 𝐴𝐵 𝑖𝑠 𝐶ℎ𝑜𝑟𝑑 𝑠𝑜 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑠𝑖𝑑𝑒 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙
∠𝐷𝐴𝐵=∠𝐵𝐻𝐴
∠𝐶=∠𝐵𝐴𝐷
𝐵𝑦 𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
Thus, a segment AHB of a circle, accepting the angle AHB (which is) equal to the given (angle) C, has been drawn on the given straight-line AB..
QED