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Inscribed Angle Theorem The measure ...

Inscribed Angle Theorem
The measure of an inscribed angle is half of the measure of the arc intercepted by it.
Given `:` In a circle with centre O, `/_BAC ` is inscribed in an arc BAC. `/_BAC` intercepts are BXC of the circle.
To prove `: m /_BAC = (1)/(2) ` m ( arc BXC ) .

Text Solution

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Proof `:` There arise three cases as shown in parts (a) , (b) , (c ) of the figure.

Case 1 `:` The centre is on the angle [as in figure (a) ]
In this case, join C to O .
`Delta OAC ` is isosceles with OA = OC .
Let `/_OAC = /_OCA = p,`
By remote interior angles theorem,
`/_ BOC = /_ OAC + /_ OCA `
`/_ BOC = p+p = 2p = 2/_ BAC `
`:. /_ BAC = (1)/(2) /_BOC `
But `/_BOC =` m ( arc BXE ) ....(Definition of measure of minor arc )
`:. /_BAC = (1)/(2) m` (arc BXC ).
Case 2 `:`
The centre is in the interior of the angle [as in figure (b) ]. In this case let D be the other endpoint of the diameter drawn through A. Let arc CMD be the one intercepted by `/_CAD`, and let arc BND be the one intercepted by `/_DAB` .
Then as proved in the first situation, we have
`/_CAD = (1)/(2) m `(arc CMD)
and `/_DAB = (1)/(2) ` m (arc BND)
`:. /_CAD + /_DAB = (1)/(2)` m (arc CMD) `+ (1)/(2) `m (arc BND)
`:. m /_ BAC = (1)/(2) [` m ( arc (CMD ) `+` m (arc BND) ]
`= (1)/(2) `m (arc BXC ) .
Case 3 `:`
The cnetre is in the exterior of the angle [ as n figure ( c ) ] . Again let D be other endpoint of the digameter drawn through A. Let arc CMD be the one intercepted by `/_CAD ` and let arc BND be the one intercepted by `/_DAB ` .
Then as proved in the first situation , we have,
`/_CAD = (1)/(2) m ( ` arc CMD )
and `/_ DAB = (1)/(2) m` (arc BND)
`:. /_CAD -/_DAB = (1)/(2) ` m (arc CMD ) `- (1)/(2) `m (arc BND)
`:. /_BAC = (1)/(2) `[ m ( arc CMD ) - m ( arc BND ) ]
`= (1)/(2) ` m (arc BXC ) .
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