Proposition 32 Raman choubay

Proposition 35 If some straight-line touches a circle, and some other straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle.

Proposition 35 Given: let some straight-line EF touch the circle ABCD at the point B let some other straight-line BD have been drawn from point B into the circle ABCD cutting it in two TO PROVE: ∠ 𝐹𝐵𝐷= ∠ 𝐵𝐴𝐷 ∠ 𝐸𝐵𝐷= ∠ 𝐷𝐶𝐵

let AD, DC, and CB have been joined. Construction let BA have been drawn from B, at right-angles to EF let AD, DC, and CB have been joined.

Proof Statements Reason 𝐵𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑟𝑒 𝐸𝑓 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑛𝑓 𝐴𝐵 ⊥𝐸𝐹 ∠𝐴𝐷𝐵= 90 0 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑠𝑒𝑚𝑖−𝑐𝑖𝑟𝑐𝑙𝑒 ∠𝐴𝐵𝐷+∠𝐵𝐴𝐷= 90 0 𝐴𝑛𝑔𝑙𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐷 ∠𝐴𝐵𝐹=∠𝐴𝐵𝐷+∠𝐵𝐴𝐷 𝐴𝑠 𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐹 𝑖𝑠 90 0 𝑏𝑦 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 ∠𝐴𝐵𝐷+∠𝐷𝐵𝐹=∠𝐴𝐵𝐷+∠𝐵𝐴𝐷 𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐹 𝑖𝑛𝑡𝑜 𝑝𝑎𝑟𝑡𝑠 Angle ABD have been subtracted from both 𝑆𝑜 ∠𝐷𝐵𝐹=∠𝐵𝐴𝐷 Thus, the remaining angle DBF is equal to the angle BAD in the alternate segment of the circle. QED

Proof Statements Reason 𝐼𝑛 𝑞𝑢𝑎𝑑𝑟𝑖𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝐴𝐵𝐶𝐷 𝑆𝑢𝑚 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑐𝑦𝑐𝑙𝑖𝑐 𝑄𝑢𝑎𝑑𝑟𝑖𝑙𝑎𝑡𝑒𝑟𝑎𝑙𝑠 ∠𝐵𝐶𝐷+∠𝐷𝐴𝐵= 180 0 𝑆𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠𝐴𝐵𝐹+∠𝐴𝐵𝐸= 180 0 𝐹𝑟𝑜𝑚 𝑎𝑏𝑜𝑣𝑒 𝑡𝑤𝑜 ∠𝐵𝐶𝐷+∠𝐷𝐴𝐵=∠𝐴𝐵𝐹+∠𝐴𝐵𝐸 ∠𝐷𝐵𝐹=∠𝐵𝐴𝐷 𝑝𝑟𝑜𝑣𝑒𝑑 ∠𝐵𝐶𝐷+∠𝐷𝐵𝐹=∠𝐴𝐵𝐹+∠𝐴𝐵𝐸 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 ∠𝐷𝐵𝐹 from both side ∠𝐵𝐶𝐷=∠𝐷𝐵𝐸 QED