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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:
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: