1. Proposition 36
Raman choubay

2. Proposition 36
If some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and one of them cuts the circle, and the other touches it, then the rectangle contained by the whole straight-line cutting the circle, and the part of it cut off outside the circle, between the point and the convex circumference, will be equal to the square on the tangent line.

3. Proposition 36
Given: Point D have been taken outside circle ABC,
And two straight-lines, DC[A] and DB, radiate from D towards circle ABC.
And let DCA cut circle ABC, and BD touch it.
TO PROVE: The rectangle contained by AD and DC is equal to the square on DB
𝑖.𝑒. 𝐴𝐷×𝐷𝐶= 𝐷𝐵 2

4. Case-1(CA through centre)
let F be the center of circle ABC
let F B have been joined.
Statements
Reason
∠𝐹𝐵𝐷= 90 0
𝐴𝑠 𝐵𝐷 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒
𝐴𝑠 𝐹 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑑 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶 𝑝𝑎𝑟𝑡 𝑜𝑓 𝐴𝐷
𝐴𝐷×𝐷𝐶+ 𝐹𝐶 2 = 𝐹𝐷 2
𝐴𝐷×𝐷𝐶+ 𝐹𝐵 2 = 𝐹𝐷 2
𝐹𝐶=𝐹𝐵 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒
𝐵𝐷 2 + 𝐹𝐵 2 = 𝐹𝐷 2
𝐴𝑠 ∆𝐹𝐵𝐷 𝑖𝑠 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝐵𝐷 2 + 𝐹𝐵 2 = 𝐴𝐷×𝐷𝐶+ 𝐹𝐵 2
𝐹𝑟𝑜𝑚 𝑎𝑏𝑜𝑣𝑒 𝑏𝑜𝑡ℎ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝐵𝐷 2 = 𝐴𝐷×𝐷𝐶
𝑆𝑞𝑢𝑎𝑟𝑒 𝑜𝑓 𝐹𝐵 𝑠𝑢𝑏𝑡𝑎𝑐𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

5. Case-2 CD not through Centre
Let E be the centre of the circle
let EF have been drawn from E, perpendicular to AC
let EB, EC, and ED have been joined E F

6. Proof Statements Reason ∠𝐸𝐵𝐷= 90 0 𝐵𝐷 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝐴𝐹=𝐹𝐶 𝐸𝐹 ⊥𝐴𝐶
∠𝐸𝐵𝐷= 90 0
𝐵𝐷 𝑖𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡
𝐴𝐹=𝐹𝐶
𝐸𝐹 ⊥𝐴𝐶
𝐴𝐷 ×𝐷𝐶+ 𝐹𝐶 2 = 𝐹𝐷 2
𝐴𝐶 𝑖𝑠 𝑝𝑎𝑟𝑡 𝑜𝑓 𝐴𝐷 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑚𝑖𝑑 𝑝𝑜𝑖𝑛𝑡
𝐴𝐷 ×𝐷𝐶+ 𝐹𝐶 2 +𝐹𝐸 2 = 𝐹𝐸 2 + 𝐹𝐷 2
𝑠𝑞𝑢𝑎𝑟𝑒 𝑜𝑓 𝐸𝐹 𝑎𝑑𝑑𝑒𝑑 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒
∆ 𝐸𝐷𝐹 𝑎𝑛𝑑 ∆ 𝐸𝐶𝐹 𝑏𝑜𝑡ℎ 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑑 𝑠𝑜 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑢𝑠 𝑡ℎ𝑒.
𝐴𝐷 ×𝐷𝐶+ 𝐸𝐶 2 = 𝐸𝐷 2
𝐴𝐷 ×𝐷𝐶+ 𝐸𝐵 2 = 𝐸𝐷 2
𝐸𝐶=𝐸𝐵 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒
𝐵𝐷 2 + 𝐸𝐵 2 = 𝐸𝐷 2
∆ 𝐸𝐵𝐷 𝑖𝑠 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑑 𝑠𝑜 𝑏𝑦 𝑃𝑦𝑡ℎ. 𝑡ℎ𝑒
𝐴𝐷 ×𝐷𝐶= 𝐵𝐷 2
𝐵𝑦 𝑅𝑒𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐸𝐷
QED