#1**+3 **

Angle OBQ = 90 ⇒ a radius meeting a tangent forrns a right angle

Angle CBO = 90 - 2x ⇒ ( m ∠OBQ - m∠ CBQ = m ∠ CBO)

Angle OCB = Angle CBO ⇒ In triangle OCB....OB = OC...so the angles opposite these sides are also equal

And in triangle DOC, DO = OC.....so angle CDO = angle DCO = x

So angle BCD = angle OCB + angle ACD = 90 - 2x + x = 90 - x

And minor arc DB = 2 * m∠ DCB = 180 - 2x [ an inscribed angle is measures1/2 of its intercepted arc ] = angle DOB

So......major arc DB = (360) - ( 180 - 2x) = 180 + 2x

And an angle external to a circle and formed by two secants, is equal to one half the difference of the intercepted arcs. This means that

y = (1/2) [ measure of major arc DB - measure of minor arc DB ]

y = (1/2) [ (180 + 2x) - (180 - 2x ) ]

y = (1/2) ( 2x + 2x)

y = (1/2) (4x)

y = 2x

CPhill
Feb 12, 2018