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  1. MHB Craftsman

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    #1


    Workings

    $\triangle ADE \cong \triangle CFE \left(AAS\right)$

    $\angle AED = \angle CEF $( vertically opposite angles )
    $\angle CFE= \angle EDA $( alternate angles )
    $AE=EC $( E midpoint )

    $ii.$ADCF is a parallelogram because diagonals bisect each other.

    Where is help needed

    How should the fact that area of $\triangle ABC$ is equal to parallelogram ADCF be proved ?

    Show that if $DE=AE$, then $\angle ADC=90^{\circ}$

  2. Perseverance
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    #2
    What is the area of triangle ADC in relation to the area of triangle ABC?

  3. MHB Craftsman

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    #3 Thread Author
    Quote Originally Posted by greg1313 View Post
    What is the area of triangle ADC in relation to the area of triangle ABC?
    The triangle ADC exactly one half of the area of the triangle ABC

    and the parallelogram ADCF is twice the area of triangle ADC and triangle ABC is twice the area of triangle ABC

    Now what remains is,

    Show that if $DE=AE$, then $\angle ADC=90^{\circ}$

    Many THanks

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    #4
    What do the measures of two opposite angles in a cyclic quadrilateral sum to?

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    #5 Thread Author
    Quote Originally Posted by greg1313 View Post
    What do the measures of two opposite angles in a cyclic quadrilateral sum to?
    180 degrees , But a cyclic quadrilateral is not to be seen here

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    #6
    Look again.

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    #7 Thread Author
    Quote Originally Posted by greg1313 View Post
    Look again.
    A little bit confused here,

    A cyclic quadrilateral is a quadrilateral whose vertices all touch the circumference of a circle.

    But here there is no circle around the figure

    Many Thanks

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    #8
    Quote Originally Posted by mathlearn View Post
    Now what remains is,

    Show that if $DE=AE$, then $\angle ADC=90^{\circ}$
    As an alternative to greg1313's suggestion, what can you say about a parallelogram whose diagonals are equal?

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    #9
    Quote Originally Posted by mathlearn View Post
    A little bit confused here,

    A cyclic quadrilateral is a quadrilateral whose vertices all touch the circumference of a circle.

    But here there is no circle around the figure

    Many Thanks
    The circle doesn't have to be there. One way of describing a cyclic quadrilateral is "A cyclic quadrilateral is a quadrilateral around which one can draw a circle that touches all four vertices". Now can you make progress with my hint (if you wish to do so)?

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    #10 Thread Author
    Quote Originally Posted by greg1313 View Post
    The circle doesn't have to be there. One way of describing a cyclic quadrilateral is "A cyclic quadrilateral is a quadrilateral around which one can draw a circle that touches all four vertices". Now can you make progress with my hint (if you wish to do so)?
    That's what was exactly missing here,

    ADFC is the cyclic quadrilateral and from there

    AFD= FDC and ADF= DFC

    ADF + FDC=ADC

    AFD+DFC=AFC

    AFD+DFC+ADF+FDC=180

    2ADF + 2FDC = 180

    dividing both sides by 2

    ADF+FDC=90

    ADC=90

    Quote Originally Posted by Opalg View Post
    As an alternative to greg1313's suggestion, what can you say about a parallelogram whose diagonals are equal?
    Taking it the alternative way,

    Both squares and rectangles have their diagonal equal & All angles of them are equal (90)

    It is a rectangle in this case as in a rectangle opposite sides are equal as in a parallelogram

    then ADC=90

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