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Math Help - Inequalities in Triangles

  1. #1
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    Inequalities in Triangles

    Firstly, you guys must be wondering how come all threads started by me are always in the 'Urgent Homework Help' section. Well, that's me.... I always get reminded of homework a li'l late.

    Secondly, Inequalities in Triangles is a topic I find difficult to understand at all. So, I'd kindly request you people to please post all the steps. Please! I understand the theorems of inequalities, however, I fail to be able to apply in the sums.


    1. In triangle ABC, if AD is a median, prove that AB + AC > 2AD

    2. In triangle ABC, AB > AC. From the greater side AB, AD is cut off equal to AC and DC is joined. Prove that -
    i) Angle ACD = 1/2 (Angle B + Angle C)
    ii) Angle BCD = 1/2 (Angle C - Angle B)


    Thanks a lot! Thanks!
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  2. #2
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    Quote Originally Posted by Ruler of Hell View Post
    ...
    Secondly, Inequalities in Triangles is a topic I find difficult to understand at all. ...
    1. In triangle ABC, if AD is a median, prove that AB + AC > 2AD
    ...
    Hello,

    to #1:

    1. Draw a sketch.
    2. Inflect the triangle ABC over the point D (not sure if this is the correct expression; see attachment)

    3. You'll get a parallelogram ABA'C. The diagonal of the parallelogram is 2*(AD)
    4. Consider the triangle ABA' with BA' = AC

    5. According to the triangle inequality
    AB + BA' > AA' = 2*(AD)
    AB + AC > 2*(AD)
    Attached Thumbnails Attached Thumbnails Inequalities in Triangles-dreiecksunglg.gif  
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  3. #3
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    Quote Originally Posted by earboth View Post
    2. Inflect the triangle
    Reflect, I think is the correct expression.

    Anyway, you have a nice proof, nicer than the once I was thinking of. The nice think about your proof is that you do not use the fact that the line is a median. Thus it works for any line inside a triangle!

    ---
    Quote Originally Posted by Ruler of Hell View Post
    2. In triangle ABC, AB > AC. From the greater side AB, AD is cut off equal to AC and DC is joined. Prove that -
    i) Angle ACD = 1/2 (Angle B + Angle C)
    ii) Angle BCD = 1/2 (Angle C - Angle B)
    You draw a picture and follow what I do.

    1)Draw triangle ABC.
    2)Mark off point D on AB.
    3)Connect C and D.
    4)Note AC = AD thus, triangle ACD is isoseles.
    5)<ADC is exterior angle for triangle DCB.
    6)Thus, <ADC = <DCB + <DBC
    7)Thus, <ACD + <ADC = (<ACD+<DCB)+<DBC
    8)Thus, <ACD + <ADC = <C + <B
    9)But <ADC = <ADC by #4.
    10)Thus, 2<ACD = <C + <B
    Q.E.D.
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  4. #4
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    @earboth: Amazing method, thanks a lot! Thanks a tonne!
    @ThePerfectHacker: The method you've shown is so much easier than what I've found. Thanks a lot!


    Thanks a lot guys, thanks a lot! Thank you!!
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  5. #5
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    Quote Originally Posted by Ruler of Hell View Post
    ...
    Secondly, Inequalities in Triangles is a topic I find difficult to understand at all. ...
    2. In triangle ABC, AB > AC. From the greater side AB, AD is cut off equal to AC and DC is joined. Prove that -
    i) Angle ACD = 1/2 (Angle B + Angle C)
    ii) Angle BCD = 1/2 (Angle C - Angle B)
    ...
    Hello,

    to #2:

    1. Draw a sketch (watch out: The angle γ is the complete angle at C)
    2. Triangle ADC is an isosceles triangle with the base CD.
    i):
    3. The red angles at the base are equal and have the value:
    <(ACD) = (180 - α)/2
    α = 180 - - γ. Now plug in this term into the previous equation:

    <(ACD) = (180 - (180 - - γ))/2 = ( + γ)/2
    ii):
    4. <(BDC) = γ - (ACD) = γ - ( + γ)/2 = γ - /2 - γ/2 = (γ - )/2
    Attached Thumbnails Attached Thumbnails Inequalities in Triangles-winkl_unglg.gif  
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