Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Geometry proof needed!!?

I need an example of a 10-15 step proof for a high school geometry class. I'll need to back it up with theorems and postulates

Update:

I am asking for an example problem! :)

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  • 1 decade ago
    Favorite Answer

    I cud only think of a 9 step proof. It's a flow proof:

    Given: ▲UVW is isosceles with vertex angle UVW

    line YV is the bisector of angle UVW

    Prove: YV is the median

    1) ▲UVW is isos. with vertex angle UVW → 2) UV ~= VW

    3) YV bisects angle UVW → 4) angle 1 ~= angle 2 }*

    5) VY ~= VY

    → 6) ▲UVY ~= ▲WVY → 7) UY ~= YW → 8) Y is the midpoint of UW → 9) YV is the median

    Theorems/Postulates:

    1) Given

    2) Definition of isosceles triangle (▲)

    3) Given

    4) Definition of angle bisector

    5) Reflexive Property

    6) Side-Angle-Side (SAS)

    7) Dei=finition of similar(~=) ▲s

    8) Definition of midpoint

    9) Definition of median

    i kno its hard to understand it when its typed and sry...i dont know how to attach the pic of the diagram...so u'll have to picture it

    * ' } ' means that i combined steps 2,4, and 5 to imply step 6

  • 1 decade ago

    ------------------------------- --------------------------

    5(4+3) - (65/5)x = 1 -----given

    20+15 -(65/5)x = 1-----distrubutive prop. a(b+c) = ab+ac

    -20 -20

    15- (65/5)x = 1-----subtraction prop. if a = b, then a-c = b-c

    +65/5 +65/5

    15 - x = 1 + 65/5-----addition prop. if a = b, then a+c = b+c

    times by 5 times by 5

    75 - 5x = 67----- divistion prop. if a = b, then a/c =b/c

    -75 -75

    -5x = -10-----subtraction prop. if a = b, then a-c = b-c

    /-5 /-5

    x = 2-----division prop. if a = b then a/c = b/c

    Source(s): hope this helps!! (pretent theres a line down the middle)
  • Anonymous
    1 decade ago

    Give the question to answer the problem

    Source(s): brain
  • CHos3n
    Lv 5
    1 decade ago

    Sorry, gave my last one any at Starbucks this morning. I do have an algebraic dilemma available...

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  • 1 decade ago

    where is the problem?

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