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Thread: 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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    $\displaystyle \angle{ACB}=90 degrees$
    EC is perpendicular to OC. AD is perpendicular to AB.
    Hence
    $\displaystyle \angle{CBA} = \angle{ECA} = \angle{EAC}$
    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

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

    so $\displaystyle \angle ECA=\angle EAC$

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

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

    Hence we can conclude that OAEC is a square

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

    $\displaystyle
    \angle DCE+\angle ECA+\angle BCA=180^o
    $

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

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

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

    AD=EC (square)

    $\displaystyle \angle AOC =\angle DEC =90^o$

    so $\displaystyle \triangle ADC \cong \triangle CED $

    ED=OC (corresponding angle)

    Since OC=EA , ED=EA
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