Let the extensions of secants AD and BC meet at point P.
Let the measure of ar DC be 2a, then m(DAC)=m(DBC)=a, since angles DAC and DBC are both inscribed angles intercepting arc DC.
m(P)=b.
Then triangles APC and BPD are similar, because they have 2 pairs of common angles.
from the similarity of APC and BPD, we can write the ratios:
[tex] \frac{AP}{BP} = \frac{PC}{PD} = \frac{AC}{BD} [/tex]
