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Math Help - plz help me with a Similarity sum...

  1. #1
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    Red face plz help me with a Similarity sum...

    In a parallelogram ABCD, P & Q are mid-points of AB and DC respectively.
    AB//PQ//DC. Join diagonal AC. AC and PQ intersect at R. Join BQ. RC and BQ intersect at S. If AS= 20cm, find,
    1. triangle CQR: triangle SCQ,
    2. triangle SCQ:triangle RSQ,
    3. quadrilateral ASQD: //gm ABCD.

    Somebody plz help me with this sum...I had tried it a million times but to no avail....
    Last edited by snigdha; February 12th 2010 at 10:57 PM.
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  2. #2
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    Parallelogram geometry

    Hello snigdha
    Quote Originally Posted by snigdha View Post
    In a parallelogram ABCD, P & Q are mid-points of AB and DC respectively.
    AB//PQ//DC. Join diagonal AC. AC and PQ intersect at R. Join BQ. RC and BQ intersect at S. If AS= 20cm, find,
    1. triangle CQR: triangle SCQ,
    2. triangle SCQ:triangle RSQ,
    3. quadrilateral ASQDarallelogram ABCD.

    Somebody plz help me with this sum....I had tried it a million times but to no avail....
    I assume that the bit I've marked in red should be AD//PQ//BC.

    Do you know how to use vectors to solve problems like this? If you do, it's quite easy. Let \vec{AB}=\vec{x} and \vec{BC} = \vec{y}. Then

    \vec{AC} = \vec{AB} + \vec{BC} = \vec{x} + \vec{y}

    \Rightarrow \vec{AS} = p(\vec{x} + \vec{y}), for some number p

    And \vec{BQ} = \vec{BC} + \vec{CQ} = \vec{y} - \tfrac{1}{2}\vec{x}, since Q is the mid-point of DC.

    \Rightarrow \vec{BS} = q(\vec{y} - \tfrac{1}{2}\vec{x}), for some number q

    But \vec{BS} = \vec{BA} + \vec{AS} = -\vec{x} +p(\vec{x} + \vec{y}) = (p-1)\vec{x} + p\vec{y}

    \Rightarrow (p-1)\vec{x} + p\vec{y} = q(\vec{y} - \tfrac{1}{2}\vec{x})

    Compare coefficients: \vec{y}: p = q; \vec{x}: p-1 = -\tfrac{1}{2}q

    \Rightarrow p = q = \tfrac{2}{3}

    So AS = \tfrac{2}{3}AC.

    Also, R is the mid-point of AC (since triangles APR and CQR are congruent), and so RC = \tfrac{1}{2}AC \Rightarrow RC : SC = 3 : 2

    \Rightarrow area triangle CQR : area triangle SCQ = 3:2, since they share the same height.

    Do you want to see if you can finish it now?

    Grandad
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  3. #3
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    Geometry Proof

    Hello snigdha

    Here's a more traditional proof that the ratio AS:SC = 2:3.

    See the attached diagram.

    PD meets AC at T. We prove first that triangles APD, CQB are congruent.

    Proof:

    AD = BC (opposite sides of a parallelogram)

    Angle A = angle C (opposite angles of a parallelogram)

    AB = DC (opposite sides of a parallelogram)

    But AP = \tfrac{1}{2} AB and CQ = \tfrac{1}{2} CD

    \therefore AP = CQ

    So the triangles APD, CQB are congruent (2 sides and included angle)

    Now we prove that triangles APT, CQS are congruent.

    Proof:

    angle APT = angle CQS (angles in congruent triangle)

    and angle PAT = angle QCS (alternate angles, AB //DC)

    and AP = QC (proven)

    \therefore triangles APT, CQS are congruent (2 angles and corresponding side)

    \therefore AT = SC

    Then consider the triangles APT and ABS. These are similar (corresponding angles equal).

    \therefore AP:PB = AT:TS

    But AP = PB \therefore AT = TS = SC

    \therefore AS : AC = 2 : 3

    Grandad
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