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How does angle of pic provided equal a/2 + 90 degrees?

  1. May 28, 2014 #1
    1. The problem statement, all variables and given/known data

    How does angle DAE equal a/2 + 90 degrees (a refers to the angle at vector A)? Please refer to the picture attached. Triangles ACE and ABD are isosceles. Side h bisects triangle ABC. Segment ED equals the perimeter of triangle ABC

    2. Relevant equations



    3. The attempt at a solution

    According to the solution, Angle DAE equals a/2 + 90 degrees. a refers to the angle at Vector A of triangle ABC. I realize the acute angles of AEC and ABD are the same in their respective triangles. Altitude h intersecting at CB makes it 90 degrees on either side, but says nothing further of the angle ( angle a) at vector A.
     

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  2. jcsd
  3. May 28, 2014 #2
    First, disregard the altitude h. It may be helpful in some way, but you don't need it. It's better to just get rid of it so that it doesn't clutter up the picture.

    Then, label the angles at vertices D and E. While you're at it, find a new label for the angle that you're calling a. The picture is already using that label for something else.

    Now label every other angle in the picture (remember, we're not paying any attention to that unneeded altitude or the angles associated with it) in terms of those three angles.

    Given what you (supposedly) know about the sum of the interior angles of a triangle, write down some equations, and see if you can't use one or more of those equations to express ∠EAD solely in terms of ∠BAC (whereas you previously had ∠EAD in terms of ∠BAC, ∠AED, and ∠EDA).

    Good luck!

    P.S. I think the word you're looking for is "vertex" rather than "vector".
     
  4. May 28, 2014 #3
    Wow. Who labeled this?! Usually the Side opposite the vertex is labeled with the same letter in lower case as the vertex. Of course you can't do that for vertex A, since there are numerous angles, but you CAN for B,C,D, & E and it wasn't done this way. Oh well. So I assume that you meant angle ∠BAC as that angle is marked with an arc at vertex A. (∠A )
    As said, ignore h (which I assume is the height, but whatever, ignore it (if you were told it was the height, then the fact that it bisects the Triangle would be a different way to figure this out, but that is a much more advanced subject).
    There are two different ways to put ∠EAD in an equation, one using the properties of triangles and the other just knowing that a total angle is the sum of its subdivisions (angles ∠EAC+∠A+∠BAD). That and the properties of isoceles trinagles is all you need.
     
  5. May 28, 2014 #4

    SammyS

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    attachment.php?attachmentid=70142&d=1401320640.png

    I agree that the labeling is somewhat confusing. However, if you consider the original triangle to be Δ ABC, then side a is opposite ∠A, side b is opposite ∠B, and side c is opposite ∠C .
     
  6. May 28, 2014 #5
    I removed the altitude in the new attached pic. I know that the internal angles of triangle ABC must equal 180 degrees. That's about as far as I could get. Perhaps I can arrive at the solution with auxillary questions. In the picture, I labeled most of the vectors with their respective angles. Notice for some angles such as <EAC and <ADB I have question marks next to the smaller case letters. My reasoning behind that is as follows. In triangle ABC, the opposite side of angle c is side c. So would that mean c? has the same angle as angle c in triangle because they have the same opposing sides? Basically I'm asking are a? b? and c? the same angles as their partners in triangle ABC?

    With that, I reasoned as follows. because c? = c, and b? = b; and because angles a + b + c = 180 degrees; a + c? + b? = 180 degrees. I thought I had something there, but from the picture, angle DAE doesn't look anything like a straight line.
     

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  7. May 29, 2014 #6

    ehild

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    See the picture. The angle of the big triangle AED at vertex A is α=β+γ+θ. And you know that α+β+y=180°.

    ehild
     

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