In the figure , two circles intersect at P and Q , and AQ is the diameter of circle APQC . THe lines APD , AQB, and DQC are straight . Prove taht
$\displaystyle
\angle ADC=\angle ABC
$
since AQ is a diameter, you have angle ACQ = 90 degrees
also, angle APQ = 90 degrees
let angles AQC = DQB = a (vertically opposite angles)
so you have angle CAQ = $\displaystyle 90^o - a$
since APD is straight line,
angles DPQ = APQ = 90 degrees
so you can prove that QD is diameter of circle PQBD
Therefore, angle QBD = 90 degrees
now you can prove that triangles ACQ and DBQ are similar,
with angles CAQ = BDQ = $\displaystyle 90^o - a$
Hence, ADBC is an inscribed quadrilateral, and so angles ADC = ABC