#1**+1 **

DCO....since CO = DO.....the angles opposite those sides are equal

So DCO = x

CBO....angle OBQ = 90° [tangent meeting a radius forms a right angle ]

So CBO = 90 - 2x

OCB .......since OB = OC, the angles opposite those sides are equal

So OCB = CBO = 90 - 2x

BCD = DCO + OCB = (x) + ( 90 - 2x) = ( 90 - x) [angle addition postulate ]

DOB is a central angle that is twice the measure of BCD = 2(90 - x) = 180 - 2x = measure of minor arc DB [ a central angle measures twice an inscribed angle intercepting the same arc ]

Note that arc DCB = 360° - minor arc DB = 360 - (180 - 2x) = 180 + 2x

[ the sum of arcs in acircle = 360° ]

Angle y = (1/2) [ difference of arc DCB - arc DOB]

[ The angle formed by the intersection of 2 tangents, 2 secants or 1 tangent and 1 secant outside the circle equals half the difference of the intercepted arcs ]

Read more: https://www.mathwarehouse.com/geometry/circle/tangents-secants-arcs-angles.php#ixzz5YI50pVkl

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

(1/2) [ 2x + 2x ] =

(1/2) [ 4x] =

2x

CPhill
Nov 29, 2018