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Triangle Proofs

  1. Mar 17, 2006 #1
    Hey,

    Check out these questions.

    1. The point P lies within the angle ABC and is equidistant from AB and BC. Prove that PB bisects the ∠ABC.

    If point P is always equidistant from AB and BC, does that mean AB and BC are the same length? If so, that would produce congruent triangles correct?

    2. A fence post 4.5 m high casts a shadow of 2.75 m. At the same time a nearby lamp post cast a shadow 13.75 m in length. Find the height of the lamp post.

    I need help getting started on this one, I think it has something to do with similar triangles.

    Thanks in advance.
     
  2. jcsd
  3. Mar 17, 2006 #2

    0rthodontist

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    No, it does not. Pick 3 points A, B, C at random and draw ABC, then draw your P and you can see that A and C can lie anywhere along the line AB and BC respectively without changing the condition on P.

    You are given that P is equidistant from AB and BC. Do you know what the distance between a point and a line is? Try drawing the line segments those distances refer to.

    This is straightforward. Draw it if you can't figure it out.
     
  4. Mar 17, 2006 #3
    For the second question, the shadow that is cast is a diagonal that extends to the base of the triangle? In that case, it's quite an easy similar triangles question.
     
  5. Mar 18, 2006 #4

    VietDao29

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    Yes, this is correct. have you worked out this problem?
    And also have you worked out the first problem? Do you know what the distance between a point and the line is?
    A hint for the first problem: You may need to use congruent triangles to tackle it.
    Can you go form here? :)
     
  6. Mar 18, 2006 #5
    No, I'm fairly confused. If point P is equidistant from AB and BC, does that mean the shortest distance from P to any point on the line AB is equal to the shortest distance from P to any point on the line BC? If I found this distance, how would that lead to congruent triangles?
     
  7. Mar 18, 2006 #6
    OK, I think I may have got it.

    Quoted from above,
    "If point P is equidistant from AB and BC, does that mean the shortest distance from P to any point on the line AB is equal to the shortest distance from P to any point on the line BC?"

    Assuming this is correct, let's call the point from P to the shortest distance on AB, D. And let's call the point from P to the shortest distance on BC, E.

    So,

    PD = PE (given - equidistant)
    PB = PB (shared)
    Angle BDP = Angle BEP (shortest distance formed by a line segment from a point to a line is 90 - perpendicular)

    Thus, Triangle BDP is congruent to Triangle BEP. (Side-Angle-Side or Hypotenuse-Side theorems).

    Therefore, AngleP BD must also equal Angle PBE and so PB bisects Angle ABC.
     
  8. Mar 18, 2006 #7
    I need some help with some more proofs with altitudes.

    1. From this image:

    [​IMG]

    BX and CY are altitudes. Prove that (AB)(AY) = (AC)(AX).

    2. From this image:

    [​IMG]

    Given that equal [in area] triangles on equal bases have equal altitudes, equal triangles having equal altitudes have equal bases, solve the following for altitude CE.
     
    Last edited: Mar 18, 2006
  9. Mar 18, 2006 #8
    clue for the 1st:
    notice that the triangle's size can be calculated in more than one way.

    clue for the 2nd:
    the triangle ABD. what kind of triangle is it? what can you tell about the length of BD?
     
  10. Mar 18, 2006 #9
    For number 1, I don't understand the clue you gave.

    For number 2, triangle ABD is right-angled. The length of BD is 30 (half of BC). How does this lead to the length of EC? I think it's because I don't really understand what they mean by the statement "Given that equal [in area] triangles on equal bases have equal altitudes, equal triangles having equal altitudes have equal bases".

    Thanks for your help so far.
     
  11. Mar 19, 2006 #10
    number 1:
    take another look at the equation (AB)(AY) = (AC)(AX).
    you can also write it like this:
    (AB)/(AC) = (AX)/(AY)
    does this reminds you of something? (hint: yes)

    number 2:
    forget about the length of BD... i was going for trigonometric approach...
    see the size of the triangle ABC. how many ways can be used to calculate this size?
     
  12. Mar 19, 2006 #11

    VietDao29

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    greytomato, I should suggest you that you should think twice before giving someone else help. And Please DO NOT confuse the OP... :grumpy:
     
    Last edited: Mar 19, 2006
  13. Mar 19, 2006 #12
    sorry for confuzing you... i'm working on my english skills :)
    (size means area)
     
  14. Mar 19, 2006 #13

    Curious3141

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    There is an easy way and a tough way to do this. Obviously you want the easy way.


    Observe triangles BAX and CAY are similar. They share a common angle A and there is a right angle in each, fixing the other angle to be equal. You can now come up with a ratio relating the lengths of corresponding sides which immediately rearranges to give the required relationship.

    The hard way involves comparing areas computed two different ways and using Pythagoras' theorem and some algebra. Forget about this unless you want to know for interest's sake.


    The easy way is to use the hint. For triangle ABC, what are two obvious ways to compute the area ? Form an equation and you get CE immediately.

    There are other harder ways using similar triangles, Pythagoras theorem and even a trigonometric approach. Don't worry about these, the question clearly wants to guide you to the easy way.
     
  15. Mar 19, 2006 #14
    OK, I got number 1. I know what you're trying to hint at for number 2. Just use the basic 1/2bh area of a triangle formula. I guess I'm just not clear on whether AD and CE are actual heights of the triangle. Are all altitudes heights?

    Also, can someone confirm the proof I had for the first question in my initial post?

    Thanks.
     
    Last edited: Mar 19, 2006
  16. Mar 19, 2006 #15

    0rthodontist

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    Your proof for the first question of your initial post is correct except for your statement that the congruence is of "side angle side" form. It is actually of "side side angle" form which doesn't necessary imply congruence, so go with the "hypotenuse and side" form instead.

    Every altitude is a height when the corresponding base is the side that the altitude is perpendicular to.
     
  17. Mar 19, 2006 #16
    Thanks Orthodontist.

    I have a couple more problems that I'm stuck on.

    [​IMG]

    Given that AB ≠ DB. Prove that AB ≠ BC using indirect reasoning.

    OK, so I guess I have to start out assuming AB = BC. This leads to saying Angle BAD = Angle BCA (ITT). But, now I have no where to go and I have not yet incorporated the AB ≠ DB in yet. Any insight?

    [​IMG]

    What's the value of triangle ABD/triangle BGC?

    For this one, if I declare a shared base, I don't have a common height.
     
    Last edited: Mar 19, 2006
  18. Mar 19, 2006 #17

    0rthodontist

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    What do the 1 and the 2 mean in the first figure?

    In the second figure, in what ways can you divide up the large triangle?
     
  19. Mar 19, 2006 #18
    I don't know what the 1 and 2 mean. That's just how it is from my textbook. My best guess would be it is used to strengthen the fact that Angles BAD and BDA are different, since AB ≠ DB.

    For 2, I have tried dividing the triangles up in many ways. If I use AB as my base, triangle ABD is 3x the base of triangle BGC but in no ways are their heights related. Similar result as if I used BC for my base.
     
  20. Mar 19, 2006 #19

    0rthodontist

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    Well, in #1 it is perfectly possible for AB to equal BC even though AB is not equal to BD. There must be additional information, perhaps given by the "1" and "2," that tells you more.

    In #2 think broader than that. How can you slice up triangle ABC?
     
  21. Mar 20, 2006 #20
    Hmm, I'll have to ask my teacher if there may have been a typo or simply the 1 and 2 mean something more significant.

    Anyways, for #2, instead of slicing up triangle ABC at all, can I not compare each of the triangles ot ABC as a whole and then to each other? So,

    triangle ABD = 1/4 triangle ABC (same height, 1/4 base)
    triangle BGC = 1/3 triangle ABC (same height, 1/3 base)

    Therefore, triangle ABD/triangle BGC = 3/4.

    Can someone please verify this method? Thanks.
     
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