Arc AXB can be
a) Semi circle
b) minor arc
c) major arc
Case 1 : Assume arc AXB is a semicircle
when Ð
ABC intercepts a semicircle the chord AB passes through the center.
Therefore m Ð
ABC = 90^{0}
(a tangent is always perpendicular to the diameter that intersects
it at the point of tangency).
m (arc BXA) = 180^{0} (arc
BXA is a semi circle)
\ m
(arc BXA) = ´
180^{0}
= 90^{0}
\ m
Ð ABC = m
(arc BXA)
Case 2 : Assume that ÐABC
intercepts a minor arc. Therefore as seen in figure 8.13 the center
O lies in the exterior of Ð
ABC.
Figure 7.13
m Ð
ABC = 90^{0}
 m Ð ABO
m Ð
ABO = 90^{0}
 m Ð
ABC  (1)
But m Ð
ABO = m Ð OAB
(as OAB is an isosceles triangle )
\
m Ð
OAB = 90^{0}
 m Ð ABC 
(2)
(1) + (2)
\ m
Ð
ABO + m Ð
OAB = 180  2 m ÐABC
Since m ÐABO
+ m Ð
OAB = 180  m Ð
BOA
180  m Ð
BOA = 180  2 m Ð
ABC
i.e. m Ð
BOA = 2 m Ð
ABC
m Ð
ABC =
m Ð BOA
m Ð
ABC =
m ( arc AXB )
Case 3 :
Figure 7.14
If Ð
ABC intercepts a major arc, the center of the circle O will lie
in the interior as ÐABC
. See figure 7.14.
Now ÐADB
intercepts a minor arc AYB.
\ m
Ð AOB = m
(arc AYB)
\ 180^{0}
 m ÐADB =
{ 360^{0}
 m (arc AXB) }
\ 180^{0}
 m ÐADB =
180^{0} 
m (arc AXB)
\ m
Ð ADB =
m (arc AXB)
If two secants intersect outside a circle half the difference in the measures of the intercepted arcs gives the angle formed by the two secants.
