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Plane Geometry Problem

  1. Jun 27, 2008 #1
    Hi everyone,
    Can any one solve the followong by plane Euclidean geometry?
    I got it by co-ordinate geometry, but couldn't get it by plane...
    >In an acute - angled triangle PQR , angle P=[tex]\pi[/tex]/6 , H is the orthocentre, and M is the midpoint of QR . On the line HM , take a point T such that HM=MT. Show that PT=2QR.
     
  2. jcsd
  3. Jul 18, 2008 #2
    Connect points T and R.
    Connect points T and Q.
    Quadrilateral HRTQ is a parallelogram because HM = MT and MR = MQ.

    Thus
    TR is perpendicular to PR, and
    TQ is perpendicular to PQ.

    Let N be the midpoint of PT.
    Connect points N and R.
    Connect points NQ.
    In right triangle PRT, NR = PT/2.
    In right triangle PQT, NQ = PT/2

    So NR = NQ = PT/2
    Thus triangle PQN is isosceles.

    Now we will show that < RNQ = 60 degrees.
    (note that < NPR = < NRP and < NPQ = < NQP)
    < RNQ = < RNT + < QNT
    = < NPR + < NRP + < NPQ + < NQP
    = 2( < NPR + < NPQ)
    = 2 < RPQ
    = 2*30
    = 60

    So triangle NPQ is equilateral
    Thus PQ = NQ

    As NQ = PT/2
    So PQ = PT/2
    PT = 2PQ

    http://www.idealmath.com
     
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