Question: Two circles intersect at point B and C. Through B, two segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. prove that angle ACPis equle to angel QCD .
The lines PBQ and ABD are straight lines which intersect at B.
The angle of intersection, ABP, is equal to the angle DBQ. ---------------------------------------- (1)
Consider the arc AP.
The angle subtended by an arc, anywhere on the circle is constant.
See here: http://www.mathopenref.com/circleinscribed.html
This means that the angle ACP, subtended by the arc AP is the same as the angle PBA (again subtended by the arc AP).
ACP = PBA ------------------------------------------------------------------------------------------------------ (2)
In the other circle, consider the arc QD.
The angle QCD, subtended by the arc QD, is the same as the angle DBQ (again subtended by the arc QD).
Since QCD = QBD and DBQ = ABP (see (1)) and ABP = ACP (see (2)) then
QED: QCD = ACP