Results 1 to 3 of 3

Math Help - Simple (should be!) Geometry Proof: Two tangents to a circle

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    168

    Simple (should be!) Geometry Proof: Two tangents to a circle

    I found this little proof in a book I was practicing out of. Does anyone have a proof for this? It must be so simple! Gaahh!

    Let y be a circle, l and m be two nonparallel lines that are tangent to y at the points P and Q, and let A be the point of intersection of l and m. Prove that:
    i) O lies on the bisector of angle(PAQ)
    ii) PA = QA
    iii) The line PQ is perpendicular to the line OA

    I know Triangle POQ is isosceles, and OP = PQ, etc etc all the things about that triangle, but I don't know how to work the point A into the mix, nor do I know anything (I don't think) about Triangle PQA. Any help here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Nacho's Avatar
    Joined
    Mar 2008
    From
    Santiago, Chile
    Posts
    135
    If H is the center of circle: <br />
\vartriangle OPA \cong OQA<br />
, (because AO=OQ (are radios), OA is side in common, and <br />
\measuredangle HPA = \measuredangle OQA = 90<br />
(the radio is perpendicular to the tangent), then for SAS, those triangules are congruent)

    for this OA is bisector of <PAQ, and PA=AQ

    Now you note that: <br />
\vartriangle OPQ<br />
is isoceles, of base PQ and how <br />
\measuredangle POA = \measuredangle QOA<br />
(congruence) <br />
 \Rightarrow PQ \bot OA<br />
for propierty of isoceles triangle
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608
    Hello, paupsers!

    Let O be the center of a circle,
    l and m be two nonparallel lines that are tangent to O at the points P and Q.
    Let A be the point of intersection of l and m.

    Prove that:

    (1) O lies on the bisector of \angle PAQ
    (2) PA = QA
    (3) PQ \perp OA
    Code:
                    *    P
               *         o
            *           /   * o
           *           /     *     o
                      /                 o
          *          /        *              o
          *       O * - - - - * - - - - - - - - - o A
          *          \        *              o
                      \                 o
           *           \     *     o
            *           \   * o
               *         o
                    *    Q

    (1) Prove: OA bisects \angle PAQ.

    OP \perp PA,\;OQ \perp QA
    . .
    A radius drawn to the point of tangency is perpencidular to the tangent.

    Hence, \Delta OPA and \Delta OQA are right triangles.

    OP = OQ
    . .
    In a circle, all radii are equal.

    OA = OA
    . .
    Reflexive property

    Hence: . \Delta OPA \cong \Delta OQA
    . .
    Side-side theorem for right triangles

    Therefore: . \angle OAP = \angle OAQ
    . .
    Corresponding parts of congruent triangles are equal.



    (2) Prove: . PA = QA
    . .
    Corresponding parts of congruent triangles are equal.



    (3) Prove: . PQ \perp OA

    OP = OQ . . .
    They are radii.

    PA = QA . . .
    proved in (1)

    Hence, OA is the perpendicular bisector of PQ.
    . .
    Two points equidistant from the ends of a line segment determine the perpendicular bisector of the line segment.

    Therefore: . PQ \perp OA

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A simple Circle Proof
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 2nd 2010, 10:24 AM
  2. Replies: 1
    Last Post: May 17th 2009, 04:12 AM
  3. Geometry proof: Two tangents to a circle
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 25th 2009, 01:33 AM
  4. Replies: 7
    Last Post: May 23rd 2008, 06:25 AM
  5. Replies: 4
    Last Post: January 18th 2007, 04:33 PM

Search Tags


/mathhelpforum @mathhelpforum