Question

I need help.ASAP.thanks

Answer 1

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 $\widehat{BD}$) is

$y+m\angle BOD+90^\circ+90^\circ=360^\circ\implies m\angle BOD=180^\circ-y$

This also means the measure of the central angle BOD that subtends major arc BCD (also the measure of the major arc BCD is)

$m\widehat{BCD}=360^\circ-(180^\circ-y)=180^\circ+y$

The inscribed angle theorem says that the measure of angle BOD is twice the measure of angle BCD, so

$m\angle BCD=90^\circ-\dfrac y2$

The interior angles of quadrilateral BCDO have sum

$\underbrace{90^\circ-2x}_{m\angle CBO}+\underbrace{180^\circ+y}_{m\widehat{BCD}}+\underbrace{x}_{m\angle CDO}+\underbrace{90^\circ-\dfrac y2}_{m\angle BCD}=360^\circ$

Simplifying this equation will give you $y=2x$.