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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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  2. #2
    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 \Delta ABC. what can you say about \angle ACB? take note the A and B are endpoints of diameter 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 \Delta ABC. what can you say about \angle ACB? take note the A and B are endpoints of diameter 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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  5. #5
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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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