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Thread: Please help me in this triangle problem

  1. #1
    Junior Member
    Joined
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    Please help me in this triangle problem

    Please help!
    Prove that

    1)<APR = <AQR (< means angle)
    2)AR and PQ bisect each other at right angles.

    Please ignore the questions written in the attached file.
    Just see the diagram and description of the diagram from there.

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Please help me in this triangle problem-untitled.jpg  
    Last edited by mr fantastic; Jul 18th 2009 at 02:48 AM. Reason: Re-formatted in lower case (that is, removed shouting)
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, saha.subham!

    Code:
                  A
                  *
                 /|\
                / | \
               /5 | 6\
            P *---+---* Q
             / \3 | 4/ \
            /   \ | /   \
           /1    \|/    2\
        B *-------*-------* C
                  R
    $\displaystyle AB \,=\,AC$
    $\displaystyle P,Q,R$ are midpoints of $\displaystyle AB, AC, BC$, respectively.

    Prove that:

    1) $\displaystyle \angle APR \,=\, \angle AQR$

    2) $\displaystyle AR$ and $\displaystyle PQ$ bisect each other at right angles.

    1) We are given: .$\displaystyle AB \,=\,AC.$
    . . So: .$\displaystyle \Delta ABC$ is isosceles: .$\displaystyle \angle 1 \,=\, \angle 2$

    Then: .$\displaystyle AP \,=\, PB \,=\, AQ \,=\, QC.$
    Also: .$\displaystyle BR \,=\,RC$

    Since $\displaystyle \Delta APQ$ is isosceles: .$\displaystyle \angle 5 \,=\,\angle 6$ .[1]


    Note: .$\displaystyle \Delta PBR \,\cong\:\Delta QCR\;\;(s.a.s.)$
    . And: .$\displaystyle PR \,=\,QR\;\text{ (corres. parts)}$
    Then: .$\displaystyle \Delta PQR$ is isosceles: .$\displaystyle \angle 3 \,=\,\angle 4$ .[2]

    Add [1] and [2]: .$\displaystyle \angle 5 + \angle 3 \;=\;\angle 6 + \angle 4$

    Therefore: .$\displaystyle \angle APR \:=\:\angle 5 + \angle 3 \;=\;\angle 6 + \angle 4 \;=\;\angle AQR$


    2) We have: .$\displaystyle \begin{array}{c}AP \,=\,AQ \\ PR \,=\,QR \end{array}$

    $\displaystyle A$ is equidistant from the endpoints of segment $\displaystyle PQ.$
    $\displaystyle R$ is equidistant from the endpoints of segment $\displaystyle PQ.$

    Therefore, $\displaystyle AR$ is the perpendicular bisector of $\displaystyle PQ.$

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