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Math Help - Geometry - Triangles - Congruency

  1. #1
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    Geometry - Triangles - Congruency

    Triangles - Congruency

    Here are 3 questions i cant solve & when asking my teacher, he couldnt solve too

    Question 1:
    Triangle ABC and Triangle DEF are such
    that AB=DE , AC=DF, AM Perpendicular to BC
    and DN Perpendicular to EF . Prove that Triangle
    ABC is Congurent to Triangle DEF
    [In the figure
    triangles are not joined. they are 2 triangles
    drawn, at side of each other & both triangles
    doesnt have any line connected.]


    Question 2:
    The Image of an object placed at point
    A before a plane mirror LM is seen at the point
    B by an observer at D. Prove that the image is
    as far behined as the object is seen in front of
    the mirror[No Figure Given]

    Question 3:
    in Triangle ABC, The bisectors of Angle B and
    Angle C meet at P. Through P, a straight line MPN is
    Drawn parallel to BC. Prove that MN = BM + CN
    [No figure Given]

    Note: Please show figures and stuff.
    Last edited by CaptainBlack; May 25th 2007 at 09:36 AM.
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  2. #2
    Member Jonboy's Avatar
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    I'm assuming these aren't problems assigned for homework otherwise your teacher would know how to work them.
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  3. #3
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    As posteed #1 is not true.
    However, if you know that AM=DN then it is true.
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  4. #4
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    Quote Originally Posted by Jonboy View Post
    I'm assuming these aren't problems assigned for homework otherwise your teacher would know how to work them.
    Yes this isnt a type of homework. these questions were given in refference books of my Grade. extra questions etc. I use the book inmy tution & my tution teacher cant solve it. i have summer holidays, so i cant ask the teacher at the moment.
    I think there is a way to solve the 1st question.

    Wont Angle BAC and Angle be equal since the lines creating them are equal?
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  5. #5
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    Hello, shaurya!

    I think I've got #3 . . .


    Question 3
    In Triangle ABC, the bisectors of angle B and angle C meet at P.
    Through P, a straight line MPN is drawn parallel to BC.
    Prove that MN = BM + CN
    Code:
                      A
                      *
                     *
                    *  *
                   *
                  *     *
                 *
                *        *
               *  *
              *     * P   *
           M *--------*---- N
            *     *     *  *
           *  *           *
        B *-----------------* C

    Since BP is the bisector of angle B: \angle PBM = \angle PBC
    Since MN \parallel BC:\;\angle MPB = \angle PBC = \angle PBM (alt-int angles)
    . . Hence: \Delta MBP is isosceles and BM = MP.

    Since CP is the bisector of angle C: \angle PCN = \angle PCB
    Since MN \parallel BC:\:\angle NPC = \angle PCB = \angle NCP
    . . Hence: \Delta NPC is isosceles and CN = PN.

    Then we have: . MN \:=\:MP + PN \:=\:BM + CN

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  6. #6
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    answers of the ones which were not answered?
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  7. #7
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    Quote Originally Posted by shaurya View Post
    Triangles - Congruency

    Here are 3 questions i cant solve & when asking my teacher, he couldnt solve too
    ...
    Question 2:
    The Image of an object placed at point
    A before a plane mirror LM is seen at the point
    B by an observer at D. Prove that the image is
    as far behined as the object is seen in front of
    the mirror[No Figure Given]
    ...
    Hello,

    1. draw a sketch of the situation.
    2. I assume that you are familiar with physical laws of reflection.

    3. You can prove that triangle LMA is congruent to triangle LMB because of ASA.

    4. Thus length of AM is equal to length of BM.
    Attached Thumbnails Attached Thumbnails Geometry - Triangles - Congruency-spiegel_asa.gif  
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