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Math Help - Prove this!

  1. #1
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    Prove this!

    In a triangle ABC, BD and CE are the medians of a triangle, meet at centroid G.
    Prove that BG=2GD.
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  2. #2
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    Centroid of a triangle

    Hello snigdha
    Quote Originally Posted by snigdha View Post
    In a triangle ABC, BD and CE are the medians of a triangle, meet at centroid G.
    Prove that BG=2GD.
    If you understand how you can use vectors in geometry, you'll find a vector proof just here.

    But if you want a more 'traditional' proof, look at the diagram I've attached.

    In the diagram, E and D are the mid-points of AB and AC respectively.

    If we consider the area of \triangle ABC when its base is AB, we see that:
    area of \triangle AEC = \tfrac12 area of \triangle ABC
    because the base ( AE) of \triangle AEC is one-half of the base ( AB) of \triangle ABC, and the height of each triangle is the same.

    Similarly, when we consider AC as the base of \triangle ABC, we get:
    \triangle ABD = \tfrac12 \triangle ABC
    Therefore:
    \triangle AEC = \triangle ABD = \tfrac12 \triangle ABC
    With the colours I've used in the diagram, this is:
    blue area + green area + yellow area = red area + yellow area + green area
    So, if we subtract the common area - the quadrilateral AEGD (yellow area + green area) - from each triangle, we get:
    \triangle BGE = \triangle CGD
    In colours:
    red area = blue area
    But we can also see that, because \triangle BGE (red) and \triangle AGE (yellow) have equal bases and the same height:
    \triangle BGE=\triangle AGE
    In colours:
    red area = yellow area
    Similarly:
    \triangle AGD=\triangle CGD
    In colours:
    green area = blue area
    Thus all four coloured triangles have the same area. Therefore
    \triangle AGB = \tfrac23\triangle ABD
    But these triangles have a common base AB. Therefore the height of \triangle AGB = \tfrac23 of the height of \triangle ABD. So, by similar triangles:
    BG = \tfrac23BD

    \Rightarrow BG = 2 GD
    Tricky, isn't it? (It's much easier to use the vector method!)

    Grandad
    Attached Thumbnails Attached Thumbnails Prove this!-untitled.jpg  
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  3. #3
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    Quote Originally Posted by snigdha View Post
    In a triangle ABC, BD and CE are the medians of a triangle, meet at centroid G.
    Prove that BG=2GD.
    hi

    first , prove that \triangle AED is similar to \triangle ABC

    Proof : \angle A is a common angle for both triangles . We also see that AB=2AE and AC=2AD . Since the ratio of 2 corresponding sides are equal and the included angles are equal , proved then .

    so \frac{AE}{AB}=\frac{ED}{BC} ----1

    Then , prove that \triangle EDG is similar to \triangle CBG .

    Proof : Since \angle AED =\angle ABC and \angle ADE=\angle ACB , ED is parallel to BC .

    hence , \angle DEG=\angle BCE , \angle EDG=\angle CBD (alternate angles) , and \angle EGD=\angle BGC (opposite angles)

    since , all corresponding angles are equal , hence proved .

    so \frac{DG}{BG}=\frac{ED}{BC} ---2

    From 1 , since \frac{AE}{AB}=\frac{1}{2} , so \frac{ED}{BC}=\frac{1}{2} , and it follows that \frac{DG}{BG}=\frac{1}{2}
    so BG=2GD
    Attached Thumbnails Attached Thumbnails Prove this!-triangle-median.bmp  
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