Results 1 to 2 of 2

Math Help - Please help me in this triangle problem

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    68

    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; July 18th 2009 at 03:48 AM. Reason: Re-formatted in lower case (that is, removed shouting)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,907
    Thanks
    765
    Hello, saha.subham!

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

    Prove that:

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

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

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

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

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


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

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

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


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

    A is equidistant from the endpoints of segment PQ.
    R is equidistant from the endpoints of segment PQ.

    Therefore, AR is the perpendicular bisector of PQ.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Triangle problem
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: September 2nd 2011, 04:40 PM
  2. triangle problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 15th 2008, 09:16 PM
  3. Triangle Problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 3rd 2008, 10:13 PM
  4. Triangle Problem
    Posted in the Geometry Forum
    Replies: 3
    Last Post: October 7th 2007, 08:05 AM
  5. Right Triangle problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 4th 2007, 09:01 PM

Search Tags


/mathhelpforum @mathhelpforum