Geometry problem (circles)

join BC and DA
let angleBAC = x, angleBDC=y, angleBCA=z, angleACD=u

in triangle BDC
sicne BD = DC = RADIUS
hence angleDBC=angleDCB
angleDBC+angleDCB+angleBCD=180 (angle sum of triangle)
hence
angleDBC= angleDCB= (180-y)/2

DA=DC(radius of circle)
angleDAC=angleDCA=u

DA=DB(radius of circle)
angleDAB=angleDBA=(x+u)

DB=DC(radius of cirlce)
angleDBC=angleDCB=(z+u)
hence (z+u)=(180-y)/2

in triangle ABC

angleABC+angleBAC+angleACB=180
(x+u+z+u)+x+z=180
2x+2(z+u)=180
2x+(180-y)=180
hence 2x = y

hence 2angle BAC = angle CDB!!!

yr question is wrong
it should be "Prove that 2 angle BAC = angle CDB"

join AD and extend the line AD to cut the circle at F
http://i32.photobucket.com/albums/d50/pzsing/circlethm.jpg

Let x=∠CDB, y=∠BAC, m=∠CAD


DA=DC (radius)
∴∠ACD=∠CAD=m (base ∠ isos. △)


∠FDC = ∠ACD + ∠CAD (ext. ∠ of △)
= 2m


∠BAD = ∠BAC+∠CAD = y+m
DA = DB (radius)
∴∠ABD = ∠BAD (base ∠ isos. △)
= y+m


∠BDF = ∠ABD + ∠BAD (ext. ∠ of △)
= 2y + 2m


However, ∠BDF = ∠CDB + ∠FDC = x + 2m
∴ 2y + 2m = x + 2m
x = 2y



i think this is simpler, right?