Both angles ABO and ADO are right angles because ABP and ADQ are tangent to circle O. The interior angles of a quadrilateral add up to 360 degrees, so the measure of angle BOD (and measure of minor arc BD, denoted [tex]\widehat{BD}[/tex]) is
[tex]y+m\angle BOD+90^\circ+90^\circ=360^\circ\implies m\angle BOD=180^\circ-y[/tex]
This also means the measure of the central angle BOD that subtends major arc BCD (also the measure of the major arc BCD is)
[tex]m\widehat{BCD}=360^\circ-(180^\circ-y)=180^\circ+y[/tex]
The inscribed angle theorem says that the measure of angle BOD is twice the measure of angle BCD, so
[tex]m\angle BCD=90^\circ-\dfrac y2[/tex]
The interior angles of quadrilateral BCDO have sum
[tex]\underbrace{90^\circ-2x}_{m\angle CBO}+\underbrace{180^\circ+y}_{m\widehat{BCD}}+\underbrace{x}_{m\angle CDO}+\underbrace{90^\circ-\dfrac y2}_{m\angle BCD}=360^\circ[/tex]
Simplifying this equation will give you [tex]y=2x[/tex].
