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December 19th, 2017, 04:15 PM  #1 
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 25 Thanks: 0 Math Focus: Calculus  How to solve this problem involving intersecting lines and angles?
The following problem might be too elementary for many of you but for me is not very obvious, and therefore I would really appreciate in the proposed answer it can be included a reworked diagram showing the why's and how's. I know drawing can be tedious specially in geometry. In this figure $BC$ is bisector of the angle $OCD$. What is the value of $\gamma$? I'm stuck at the value of $\gamma$ as it is in a position from where I cannot relate if with any other angle. What are the identities or lemmas required to solve this thing?. 
December 19th, 2017, 04:31 PM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 1,781 Thanks: 919  Quote:
$\angle BCD = \dfrac 1 2 \angle OCD = 10^\circ$ $\angle BDC = 25^\circ$ $180^\circ = \angle BDC + \angle BCD + (180^\circ\gamma)$ $180^\circ = 25^\circ + 10^\circ + 180^\circ  \gamma$ $\gamma = 25^\circ + 10^\circ =35^\circ$ Last edited by romsek; December 19th, 2017 at 05:13 PM.  
December 19th, 2017, 05:13 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,689 Thanks: 1522 
Here's an accurate diagram. TriangleBCD.JPG By the exterior angle theorem, $\gamma = \angle\text{BDC} + \angle\text{BCD} = 25^\circ + 10^\circ = 35^\circ$. 

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angles, bisector, intersecting, involving, lines, problem, solve 
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