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# proof of circle theorems

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https://vle.mathswatch.co.uk/images/questions/question15392.png

Feb 11, 2018

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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   Feb 12, 2018
edited by CPhill  Feb 12, 2018