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Thread: Complex geometry

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    Complex geometry

    AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.

    Is it even possible to prove it? I mean I think BC is less that half of BD...
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ice_syncer View Post
    AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.

    Is it even possible to prove it? I mean I think BC is less that half of BD...
    the picture is not accurate but it is something like that..

    now, consider triangle $\displaystyle \Delta ABC$. what can you say about $\displaystyle \angle ACB$? take note the $\displaystyle A$ and $\displaystyle B$ are endpoints of diameter $\displaystyle AB$
    Attached Thumbnails Attached Thumbnails Complex geometry-rep.gif  
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    Quote Originally Posted by kalagota View Post
    the picture is not accurate but it is something like that..

    now, consider triangle $\displaystyle \Delta ABC$. what can you say about $\displaystyle \angle ACB$? take note the $\displaystyle A$ and $\displaystyle B$ are endpoints of diameter $\displaystyle AB$
    Sorry I didn't understand..
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    Quote Originally Posted by ice_syncer View Post
    Sorry I didn't understand..
    OK got it.
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    Quote Originally Posted by ice_syncer View Post
    AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.

    Is it even possible to prove it? I mean I think BC is less that half of BD...
    angle BAC = 30 degrees ....given.

    angle ACB = 90 degrees....because it is subtended by a diameter, by a semi-circle.

    So, angle ABC = 180 -30 -90 = 60 degrees.

    angle ABD is a straight angle, so it is 180 degrees.
    And so, angle CBD = 180 -angle ABC = 180 -60 = 120 degrees.

    Minor arc BC = 2(angle BAC) = 2(30) = 60 degrees....an inscribed angle is half of the intercepted arc in measure.
    So, angle BCD = (1/2)(60) = 30 degrees....a tangent-chord angle is half of the intercepted arc in measure.
    So, angle BDC = 180 -120 -30 = 30 degrees.

    Hence, triangle CBD is an isosceles triangle, whose equal sides are BC and BD.

    Therefore, BC = BD.
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