Let the vertex of an `/_ABC` be located outside a circle and let the sides of the angle intersect equal chords `AD` and `CE` with the circle. Prove that `/_ ABC` is equal to half of the difference of the angles subtended by the chords `AC` and `DE` at the centre.

Explanation

Let `O` be the centre of the given circle.
Let `ADB` and `CEB` be the two chords of the circle such that `AD` `=` `CE`.
Join `OA, OC, OD , OE` and `DE`.
Then , we have to prove that `/_ABC = (1)/(2) ( /_ AOC - /_ DOE )` .
In `Delta OAD` and `Delta OCE`, we have
`OA` `=` `OC` [radii of the same circle ]
`OD` `=` `OE` [radii of the same circle ]
`AD` `=` `CE` [given]
`:. Delta OAD ~= Delta OCE `[ by `SSS` -congruence ].
So, `/_ AOD = /_ COE , OA = OC ` and OD `=` OE.
`:. /_ OAD = /_ ODA = /_ OCE = /_ OCA = y^(@) ` ( say )
`:. /_ AOC = (180^(@) - 2y^(@))` . ...(i)
In `DeltaODE`, we have
`OD = OE implies /_ ODE = /_ OED = z^(@) ` ( say ).
`:. /_ ODE = (180^(@) - 2z^(@))` ....(ii)
Clearly, ACED is a cyclic quadrilateral.
`:. /_ C + /_ D = 180^(@) implies ( x^(@) + y^(@)) + (x^(@) + z^(@)) = 180^(@)`
`implies 2x^(@) + y^(@) +z^(@) = 180^(@)` ....(iii)
Now, in `Delta BDE`, we have
`/_ BDE + /_BED + /_ DBE = 180^(@)`
`implies [180^(@) - (x^(@) +z^(@))]+[180^(@)-(x^(@)+z^(@))]+/_DBE = 180^(@)`
`implies /_DBE = 2x^(@) +2z^(@)-180^(@)`
`=180^(@) - (y^(@)+z^(@))+2z^(@)-180^(@) ` [using (iii) ]
`implies /_DBE = z^(@) -y^(@)`
`implies /_ABC = z^(@) -y^(@) ` .......(iv) `[ :' /_ DBE =/_ABC ]`
From (i) and (ii) ,we get
`( /_ AOC - /_ DOE ) = (180^(@) -2y^(@)) - ( 180^(@)-2z^(@))= 2 (z^(@)-y^(@))`
`implies (1)/(2) (/_AOC - /_DOE ) = z^(@) -y^(@)` ....(v)
From (iv) and (v) , we have
`(1)/(2) (/_ AOC - /_ DOE) = /_ ABC`