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Math Help - basic geometry proof

  1. #1
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    Question basic geometry proof

    stuck on how to go about with this problem




    so I guess I need to show PQR or PQS is 90 deg.

    any ideas?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hello View Post
    stuck on how to go about with this problem




    so I guess I need to show PQR or PQS is 90 deg.

    any ideas?
    yes, i have an idea. the problem is a lot easier than you think. all you need to know is that angles in a triangle sum to 180 degrees. (oh yeah, and you need to know about similar triangles as well, corresponding sides and angles are proportional. indeed, if hey share a corresponding side, then all corresponding sides and angles are exactly equal)
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  3. #3
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    In triangles PQR and PSQ,
    \angle PRQ = \angle PSQ (data)
    RQ = QS (Q is the midpoint of RS)
    PQ is common
    Hence, triangles PQR and PSQ are congruent (SAS)
    \angle PQR = \angle PQS (corresponding angles in congruent triangles are equal)
    180 degrees = \angle PQR + \angle PQS (angle sum of straight \angle RQS equals 180 degrees)
    Therefore, \angle PQR = \angle PQS = 90 degrees
    Hence, PQ is perpendicular to RS.
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