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PROVE: The altitude of a triangle are concurrent.

  1. Sep 9, 2006 #1
    Hey,
    this proof has been annoying me all day. i don't understand what its asking do and have no idea how to solve it. The proof asks:

    Prove that the altitudes of a traingle are concurrent.
    Here is the diagram for it:

    [​IMG]

    [tex]
    \begin{array}{l}
    \overrightarrow {FA} = {\bf{a}} \\
    \overrightarrow {FC} = {\bf{c}} \\
    \overrightarrow {FB} = {\bf{b}} \\
    \end{array}
    [/tex]

    Im also told to use the fact that

    [tex]{\bf{a}} \cdot \overrightarrow {BC} = 0[/tex]

    and

    [tex]{\bf{b}} \cdot \overrightarrow {AC} = 0[/tex]

    to prove that [tex]\overrightarrow {CF}[/tex] is perpendicualr to [tex]\overrightarrow {AB}[/tex]

    thanks to anyone who understands what this is asking me to do and post back helpful infomation,
    Pavadrin
     
    Last edited: Sep 9, 2006
  2. jcsd
  3. Sep 9, 2006 #2

    StatusX

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    Homework Helper

    An altitude of a triangle is a line that passes through one of the points and is perpendicular to the adjacent side. There are three such lines, and this problem asks you to show that they are always concurrent, which just means that they intersect at a point (note that any two non-parallel lines intersect at a point, but it is not always the case that three lines will). The hint they have given you is to show that if F is the point of intersection of two of the altitudes (which, again, must exist), then the line passing through C and F is perpencidular to AB, which means this line is the third altitude, and so all three altitudes pass through F.
     
  4. Sep 9, 2006 #3
    thank you for your reply explaining what this geometric proof is regarding
     
  5. Sep 10, 2006 #4
    here i will atempt to prove that [tex]\overrightarrow {CF}[/tex] is perpendicular to [tex]\overrightarrow {AB}[/tex]

    ____________________​

    [tex]
    \begin{array}{c}
    {\rm{prove }}\overrightarrow {CF} \bot \overrightarrow {AB} \\
    {\bf{a}} \cdot \overrightarrow {BC} = 0 \\
    {\bf{a}} \cdot \left( {\overrightarrow {BF} + \overrightarrow {FC} } \right) = 0 \\
    {\bf{a}} \cdot \left( { - {\bf{b}} + {\bf{c}}} \right) = 0 \\
    - {\bf{a}} \cdot {\bf{b}} + {\bf{a}} \cdot {\bf{c}} = 0 \\
    {\rm{therefore }}{\bf{a}} \cdot {\bf{b}} = {\bf{a}} \cdot {\bf{c}} \\
    {\bf{b}} \cdot \left( {\overrightarrow {AC} } \right) = 0 \\
    {\bf{b}} \cdot \left( {\overrightarrow {AF} + \overrightarrow {FC} } \right) = 0 \\
    {\bf{b}} \cdot \left( { - {\bf{a}} + {\bf{c}}} \right) = 0 \\
    - {\bf{a}} \cdot {\bf{b}} + {\bf{b}} \cdot {\bf{c}} = 0{\rm{therefore }}{\bf{a}} \cdot {\bf{b}} = {\bf{b}} \cdot {\bf{c}} \\
    {\rm{therefore }}{\bf{a}} \cdot {\bf{b}} = {\bf{a}} \cdot {\bf{c}} = {\bf{b}} \cdot {\bf{c}} \\
    \overrightarrow {CF} = - {\bf{c}} \\
    \overrightarrow {AB} = \overrightarrow {AF} + \overrightarrow {FB} \\
    = - {\bf{a}} + {\bf{b}} \\
    \overrightarrow {CF} \cdot \overrightarrow {AB} = - {\bf{c}} \cdot \left( { - {\bf{a}} + {\bf{b}}} \right) \\
    = {\bf{a}} \cdot {\bf{c}} - {\bf{b}} \cdot {\bf{c}} \\
    {\rm{therefore since }}{\bf{a}} \cdot {\bf{c}} = {\bf{b}} \cdot {\bf{c}} \Rightarrow \overrightarrow {CF} \cdot \overrightarrow {AB} = 0 \\
    {\rm{therefore }}\overrightarrow {CF} \bot \overrightarrow {AB} \\
    \end{array}
    [/tex]

    ____________________​

    is what i have done correct or have i taken the wrong assumptions? thanks,
    Pavadrin
     
  6. Jul 27, 2011 #5
    this prove is not correct.....
     
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