Question

Theorem of angle between tangent and secant
If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc .

Given `:` Let O be the centre of the circle. Line DBC is tangent to the circle at point B. Seg BA is a chord of the circle. Point X of the circle is on C side of line BA and point Y of the circle is on D side of line BA.
To prove `:` m `/_ ABC = (1)/(2) m (arc AXB)`.

Answer 1

Proof `:` Considering `/_ABC ` and the arc AXB intercepted by it, we see from the figures that arc AXB may be a semicircle, may be a minor arc or may be a major arc. We deal with these cases separately .
(a) When `/_ ABC` intercepts a semicircle , [as in figure (a) ] the chord BA passes through the centre.
In that case, `/_ABC = 90^(@)` ....(Tangent theorem )
Also since m ( arc BXA `= 180^(@)`
`:. (1)/(2) m ` (arc BXA ) ` = (1)/(2) xx180^(@) = 90^(@)`
`:.` m ` /_ABC = (1)/(2) m ` (arc BXA )
(b) When `/_ ABC ` intercepts a minor arc, [ as in figure (b) ] the centre lies in the exterior of `/_ ABC `.
We have `, m /_ABC = 90^(@) - m /_ABO`
`:. m /_ ABO = 90^(@) - m /_ ABC `
`:. m /_BAO = 90^(@) - m /_ ABC ` ....(`Delta ABC ` is isosceles )
`m /_ABO + m /_BAO = 180^(@) - 2m/_ABC `
`:. 180^(@) - m /_ BOA = 180^(@) - 2m /_ABC `
`:. 180^(@) - m /_BOA = 180^(@) - 2m/_ABC `
`:. m /_BOA = 2m/_ABC `
But `m/_BOA = m (arc AXB ) `
`:. m/_ABC = (1)/(2) m ` (arc AXB)
(c) When `/_ABC ` intercepts a major arc [as in figure (c ) the centre lies in the interior of `/_ABC ` .
We then have `/_DBA ` intercepting a minor arc BYA .
`:. m/_DBA = (1)/(2) ` m (arc BYA )
`:. 180^(@) - m/_ABC = (1)/(2) [ 360^(@) - m ` (arc AXB ) ]
`= 180^(@) - (1)/(2)` m (arc AXB )
`:. m /_ABC = (1)/(2) ` m (arc AXB )
Thus, in every case we have shown that `m/_ABC = (1)/(2)` m (arc AXB ) .