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Math Help - difficult geometry questions--please help

  1. #1
    shosho
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    difficult geometry questions--please help

    hi, i have 2 hard geometry questions that i just cannot do. any help would be great

    1) In a pentagon ABCDE, AB=AE, BC=ED, and BD=EC Prove that AD=AC

    2)In a quadrilateral ABCD, the diagonal AC bisects <BAD and AB=BC=CD. Prove that the other diagonal BD, bisects <ADC

    thanx
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  2. #2
    Senior Member tukeywilliams's Avatar
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    1. You have to show that m<ADC = m < ACD ( ADC is an isosceles triangle)
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  3. #3
    MHF Contributor
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    Quote Originally Posted by shosho View Post
    hi, i have 2 hard geometry questions that i just cannot do. any help would be great

    1) In a pentagon ABCDE, AB=AE, BC=ED, and BD=EC Prove that AD=AC

    2)In a quadrilateral ABCD, the diagonal AC bisects <BAD and AB=BC=CD. Prove that the other diagonal BD, bisects <ADC

    thanx
    Nice problems. Good for practice.
    Let me show it not in statements-reasons form.

    1) In a pentagon ABCDE, AB=AE, BC=ED, and BD=EC Prove that AD=AC
    You have to sketch the figure on paper. I do not know how to sketch figures on the board.

    The idea is to show that triangle ABC is congruent to riangle AED so that AD=AC.

    First, triangle BCD is congruent to triangle EDC by SSS, because
    BC=ED, BD=EC as givens, and CD=DC for the one and the same side.
    So, angle CBD = angle DEC. -------------(i)

    Draw line segment BE.
    Triangles CBE and DEB are conguent, by SSS again.
    BC=ED, BD=EC as givens, and BE=BE for the one and the same side.
    So, angle DBE = angle CEB. -------------(ii)

    In triangle ABE, since AB=AE, then the triangle is isosceles, and so,
    angle EBA = angle BEA ------------------(iii)

    So, using (i),(ii) and (iii), anlge ABC = angle AED.
    Hence, triangles ABC and AED are congruent, by SAS, because
    AB=AE, angle ABC = angle AED, and BC=ED.

    Therefore, AD = AC.

    ==================================
    2)In a quadrilateral ABCD, the diagonal AC bisects <BAD and AB=BC=CD. Prove that the other diagonal BD, bisects <ADC

    This one is tougher. Took me many tries. The SSA (or ASS) was a block.

    Draw the figure on paper. Draw the diagonal AC.

    AB=BC, so triangle ABCis isosceles, and so angle BAC = angle BCA.

    angle BAC = angle DAC -------given, or they are half each of angle BAD.

    So angle DAC = angle BAC = angle BCA.
    Hence, side AD is parallel to side BC. -----reverse of alternate interior angles of parallel lines (and a transversal line, diagonal AC here) are equal.

    So, using diagonal BD as the transversal,
    angle ADB = angle CBD ----alternate interior angles are equal. ----(a)

    But in isosceles triangle BCD, since BC=CD,
    angle CBD = angle CDB -----------------------(b)

    So, using (a) and (b),
    angle ADB = angle CDB.

    Therefore, diagonal BD bisects angle ADC.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by tukeywilliams View Post
    1. You have to show that m<ADC = m < ACD ( ADC is an isosceles triangle)
    Construct the perpendicular bisector of CD this passes through
    the point of intersection of EC and BD it also passes through A.
    (sorry this does not quite look right in the attachment, but the extra line
    is the perpendicular bisector that I require, can you see why it passes through
    A and the point of intersection of EC and BD?)

    The triangles formed by the bisector the two halves of CD and AC
    and AD are right triangles with two equal sides and so congruent,
    and as AC and AD are corresponding sides in these triangles they
    are congrent.

    RonL
    Attached Thumbnails Attached Thumbnails difficult geometry questions--please help-gash.jpg  
    Last edited by CaptainBlack; August 4th 2007 at 12:57 AM.
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  5. #5
    shosho
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    thanx for your help

    but captainblack.. i like how your method is short and concise but imnot sure how i can prove that the perpendicular goes through the point of intersection?? could you please explain?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by shosho View Post
    thanx for your help

    but captainblack.. i like how your method is short and concise but imnot sure how i can prove that the perpendicular goes through the point of intersection?? could you please explain?
    consider the bisector of CD that passes through the point of intersection
    and label the mid point of CD M and the point of intersection I for
    convenience.

    Then CI is congruent to DI as triangles DBC and CED are congruent and CD
    is a common side (angles BDC and ECD are corresponding angles in these
    triangles and so equal which makes triangle DIC isosceles). So triangles CIM
    and DIM are congruent by SSA. But angles CMI and DMI are corresponding
    angles and so congruent, but they are also supplementary angles and so sum
    to 180 degrees, so each or them is 90 degrees.

    Hence the line through I and M is a perp bisector of CD.

    (alternatively it is obvious by symmetry that the bisector of CD passing
    through M is a perp bisector)

    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by shosho View Post
    thanx for your help

    but captainblack.. i like how your method is short and concise but imnot sure how i can prove that the perpendicular goes through the point of intersection?? could you please explain?
    An even shorter demonstration follows from the observation that the
    construction has reflection symmetry about MI produced (the bisector of CD through I).

    Then as AC is the image of AD under reflection in the sysmmetry line they
    have the same length.

    RonL
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