prove that angle ACPis equle to angel QCD

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.

Answer 1

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

Answer 2

????????? wot angels mite yu make nois bout ????????

Wen 2 serkels hit eech oterh, & yu hav 2 hit points (not tangent)

draw strate line tween 2 hit points

Then draw line tween 2 senters

2 lines 90 deg tu eech other & form 4 rite triangess...2 sets av 2 identikal triankels

So yu hav a lotta angels = korresponding angel in its twin

Won triangel hav 2 sides= radius av serkel 1 & anuther hav 2 sides=radius av serkel 2

Then other side=konnon side with its twin