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# I have a hard circle theorem problem:

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I have a hard circle theorem problem:

The question is here: http://imgur.com/539N56S

Specifically, I want to know why part b) is 60 degrees.

Thanks for any help in advance.

bqrs01  May 1, 2017
edited by bqrs01  May 1, 2017
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### 1+0 Answers

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a.

BAD  = 70°

Since BDA  has its endpoints on a diameter, it measures 90°

And ABD  =   180 - 70  - 90  =  20°

Since BD is a trasnsversal cutting two parallel lines....BDC  =  ABD.....therefore, BDC  = 20°

b.

Note..... BCD  is supplemental to BAD    =  110°

And angle DBC   =  180  -  BDC - DCB =  180  - 20  - 110  =  50°

And minor arc DC  =  2DBC  =  100°

And angle CDT  =  (1/2) minor arc DC  = 50°

And by the experior angle theorem, angle BCD  =  angle CDT  + angle CTD  ...so..

BCD  = CDT  + CTD

110  =  50   +  angle CTD    .....subtract 50 from both sides

60°  = angle CTD  = angle BTD

CPhill  May 1, 2017

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