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Math Help - Circle Geometry problem

  1. #1
    RAz
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    Question Circle Geometry problem

    Hi! I'm not sure where to post this problem, but I guess under "Geometry" will do for now, right?

    Well anyway, I'm having a little problem with circle geometry. It's my first time solving problems like the one I'm about to explain (I do, however, know the basics). Right, here I go:

    Question: In a circle with center O, two chords AC and BD intersect at P. Show that LAPB = 1/2 (LAOB + LCOD).

    (The 'L' is the symbol for angle, it should be in italics)


    So the two normal lines, AC and BD, intersect at P (which could be anywhere in the circle). The middle of the circle, O, is the black squiggle. The dotted lines are the 'extra/invisible' lines which also make up triangles. And now...I have no idea how to solve it.

    Any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by RAz View Post
    Hi! I'm not sure where to post this problem, but I guess under "Geometry" will do for now, right?

    Well anyway, I'm having a little problem with circle geometry. It's my first time solving problems like the one I'm about to explain (I do, however, know the basics). Right, here I go:

    Question: In a circle with center O, two chords AC and BD intersect at P. Show that LAPB = 1/2 (LAOB + LCOD).

    (The 'L' is the symbol for angle, it should be in italics)


    So the two normal lines, AC and BD, intersect at P (which could be anywhere in the circle). The middle of the circle, O, is the black squiggle. The dotted lines are the 'extra/invisible' lines which also make up triangles. And now...I have no idea how to solve it.

    Any help would be greatly appreciated!
    In the given figure, draw chords BC and AD.

    For the minor arc AB:
    <AOB = arc AB in degrees.
    and <ACB = (1/2) arc AB in degrees
    So, <ACB = (1/2) <AOB. -------------**

    For minor arc CD:
    <COD = arc CD in dgrees
    and, <CBD = (1/2) arc CD in degrees,
    So, <CBD = (1/2) <COD. -------**

    In triangle BPC:
    <BPC = 180 -(<ACB + <CBD)

    In straight angle APC:
    <APB = 180 - <BPC
    So, <APB = 180 -[180 -(<ACB +<CBD)]
    <APB = <ACB +<CBD
    And so,
    <APB = (1/2) <AOB +[(1/2) <COD)]
    <APB = (1/2)[<AOB + <COD] --------------proven.
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  3. #3
    RAz
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    Quote Originally Posted by ticbol View Post
    In the given figure, draw chords BC and AD.

    For the minor arc AB:
    <AOB = arc AB in degrees.
    and <ACB = (1/2) arc AB in degrees
    So, <ACB = (1/2) <AOB. -------------**

    For minor arc CD:
    <COD = arc CD in dgrees
    and, <CBD = (1/2) arc CD in degrees,
    So, <CBD = (1/2) <COD. -------**

    In triangle BPC:
    <BPC = 180 -(<ACB + <CBD)

    In straight angle APC:
    <APB = 180 - <BPC
    So, <APB = 180 -[180 -(<ACB +<CBD)]
    <APB = <ACB +<CBD
    And so,
    <APB = (1/2) <AOB +[(1/2) <COD)]
    <APB = (1/2)[<AOB + <COD] --------------proven.
    Thank you so much for replying ticbol! There is one thing I don't quite understand though;

    "For the minor arc AB:
    <AOB = arc AB in degrees"

    What do you mean by it? I know the arc is the little curved line between A and B, but how does the <AOB equal the arc?

    Sorry for the second question.
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  4. #4
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    Quote Originally Posted by RAz View Post
    Thank you so much for replying ticbol! There is one thing I don't quite understand though;

    "For the minor arc AB:
    <AOB = arc AB in degrees"

    What do you mean by it? I know the arc is the little curved line between A and B, but how does the <AOB equal the arc?

    Sorry for the second question.
    No need to feel sorry. We want you to ask if our answers/replies are not yet clear to you.

    I assumed you know that in a circle, the central angle is equal in measure to the intercepted arc, in degrees.
    If not yet, please search the internet, Google, for example, for "angles in circles". You will see there why.
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