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
) is
![y+m\angle BOD+90^\circ+90^\circ=360^\circ\implies m\angle BOD=180^\circ-y](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s52v84u66iac674op03zsrqoezgudu70g8.png)
This also means the measure of the central angle BOD that subtends major arc BCD (also the measure of the major arc BCD is)
![m\widehat{BCD}=360^\circ-(180^\circ-y)=180^\circ+y](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pwmvackh4x6v6nnbn48rlzd49d141txihs.png)
The inscribed angle theorem says that the measure of angle BOD is twice the measure of angle BCD, so
![m\angle BCD=90^\circ-\frac y2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u224i4tjhqhkhfoo9t9bv6m4nj5fkslmqr.png)
The interior angles of quadrilateral BCDO have sum
![\underbrace{90^\circ-2x}_(m\angle CBO)+\underbrace{180^\circ+y}_{m\widehat{BCD}}+\underbrace{x}_(m\angle CDO)+\underbrace{90^\circ-\frac y2}_(m\angle BCD)=360^\circ](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2b3g0qkx24scbauffdvy1slb1an0luqde2.png)
Simplifying this equation will give you
.