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Math Help - Plane Geometry

  1. #1
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    Plane Geometry

    Plane Geometry-maths.bmp

    The diagram shows a circle, centre O, with diameter AB. The point C lies on the circle. The tangent to the circle at A meets BC extended at D. The tangent to the circle at C meets the line AD at E.
    How to prove that E is the mid-point of AD?

    Thanks.
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  2. #2
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    \angle{ACB}=90 degrees
    EC is perpendicular to OC. AD is perpendicular to AB.
    Hence
    \angle{CBA} = \angle{ECA} = \angle{EAC}
    So EA = EC.
    Similarly you can prove that EP = EC.
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  3. #3
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    Quote Originally Posted by fantasylo View Post
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    The diagram shows a circle, centre O, with diameter AB. The point C lies on the circle. The tangent to the circle at A meets BC extended at D. The tangent to the circle at C meets the line AD at E.
    How to prove that E is the mid-point of AD?

    Thanks.
    hi

    \angle ECA=\angle CBA -- 1
    \angle EAC=\angle CBA (alternate segments)

    so \angle ECA=\angle EAC

    and EA=EC (equal sides of an isosceles triangle)

    \angle OAE=\angle OCE=90^o (tangents at A and C)

    Hence we can conclude that OAEC is a square

    \angle BCA=90^o (angle in a semicircle)

     <br />
\angle DCE+\angle ECA+\angle BCA=180^o<br />

    \angle DCE+\angle ECA=90^o (angle of a line)

    \angle OAC+\angle BCA+\angle CBA =180^o

    From 1 , we see that \angle CAB=\angle ECD

    AD=EC (square)

    \angle AOC =\angle DEC =90^o

    so \triangle ADC \cong \triangle CED

    ED=OC (corresponding angle)

    Since OC=EA , ED=EA
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