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Math Help - In desperate need of help! 3 Problems

  1. #1
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    In desperate need of help! 3 Problems

    My Discrete exam is coming up in two days and I just realized that I didn't copy down solutions for a really hard proofs test that I had. These are three questions that I had no idea how to do on my test. Please help!

    1. Given: ABCD is a parallelogram with equilaterals ABF and ADE drawn on sides AB and AD, respectively. Prove that triangle FCE is equilateral.

    2. P is any point on rectangle ABCD. Prove that PA^2 + PC^2 = PB^2 + PD^2

    3. In triangle ABC, altitude BE is extended to G so that EG = the measure of altitude CF. A line through G and parallel to AC meets BA extended at H, as shown in the figure. Prove that AH = AC.

    Diagrams are attached (in order)! Thanks alot!
    Attached Thumbnails Attached Thumbnails In desperate need of help! 3 Problems-img_0611.jpg   In desperate need of help! 3 Problems-img_0614.jpg   In desperate need of help! 3 Problems-img_0615.jpg  
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  2. #2
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    Krizalid's Avatar
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    For the first one

    \triangle{AEF}\cong\triangle{DEC}\implies\overline  {EF}=\overline{CE}, but \measuredangle~AEF=\measuredangle~CED\implies\meas  uredangle~CEF=60^\circ\,\therefore\,\triangle{FCE}  \text{ is equilateral}\,\blacksquare
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    For the second one

    Draw a line through P such that cuts \overline{AB} and \overline{CD} orthogonally.

    Then play with Pythagorean Theorem.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Finally, for the third one, I'll give you a

    Hint

    Note that \triangle{ABE}\sim\triangle{HBG} and \triangle{ACF}\sim\triangle{ABE}

    Make a fusion with this equalities to get what you want.

    See ya
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