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December 19th, 2017, 05:15 PM   #1
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Question 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?.
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December 19th, 2017, 05:31 PM   #2
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Quote:
Originally Posted by Chemist116 View Post
In this figure $BC$ is bisector of the angle $OCD$. What is the value of $\gamma$?

$\angle OCD = 180^\circ - 160^\circ = 20^\circ$

$\angle BCD = \dfrac 1 2 \angle OCD = 10^\circ$

$\angle BDC = 25^\circ$

$180^\circ = \angle BDC + \angle BCD + (180^\circ-\gamma)$

$180^\circ = 25^\circ + 10^\circ + 180^\circ - \gamma$

$\gamma = 25^\circ + 10^\circ =35^\circ$
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Last edited by romsek; December 19th, 2017 at 06:13 PM.
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December 19th, 2017, 06:13 PM   #3
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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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