If G is the centroid of triangle ABC, Prove that Area of triangle GAB = 1/3 Area of triangle ABC.

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Given `Delta ABC` with centroid G, prove that `"Area"Delta GAB=1/3"Area"Delta ABC` .

The centroid of a triangle is the intersection of the medians. A median is the segment drawn from a vertex of the triangle to the midpoint of the side opposite the vertex.

Let M be the midpoint of `bar(AC)` , N the midpoint of `bar(BC)` and O the midpoint of `bar(AB)` . Let P be the intersection of `bar(MN),bar(CO)` .

(1)The proof is easy if you know that the centroid divides each median into 2 segments in the ratio of 2:1 with the segment from the vertex longer than the segment to the side. Assuming you know this, then use the formula for the area of a triangle `"Area"=1/2bh` and let the base be `bar(AB)` . The the area of `Delta ABC=1/2(AB)(CO)` and the area of `Delta GAB=1/2(AB)(GO)` where `GO=1/3CO` . Then the proof is complete.

(2) If you do not know this, we can use similar triangles:

Recall that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half the length of the third side.

So `Delta CMN ` is similar to `Delta CAB` by AA similarity. Also `Delta GAB` is similar to `Delta GMN` by AA similarity. Both triangles have sides in ratio 2:1.

Then GM=2AN, GN=2BM, GO=2GP

GO+GP+CP=CO But `CP=1/2CO` so

GO+GP=1/2CO But GO=2GP so

3GP=1/2CO ==> GP=1/6CO ==>GO=1/3CO

Now `"Area" Delta CAB=1/2(AB)(CO)` and `"Area" Delta GAB=1/2(AB)(GO)=1/2(AB)(1/3CO)=1/3"Area"Delta CAB` as required.

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Let's assume that vertices of triangle have following coordinates:

`A=(x_A,y_A)`, `B=(x_B,Y_B)`, `C=(x_C,y_C)`.

Since centroid is, in a way, a generalisation of arithmetic mean it's coordinates are:


If we have coordinates of triangles vertices we can calculate its area by following formula:

`T_(ABC)=1/2|(x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y_A)|` that is area of triangle ABC.

Let's calculate area of triangle ABG:


` ` `(x_A-x_B)(1/3y_A+1/3y_B+1/3y_C-y_A)|=`





` 1/3[1/2|(x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y_A)|] =1/3T_(ABC)`

If you look at the begining of this calculation and at the end you'll see that this proves that area of triangle ABG is 1/3 of the area of triangle ABC. 

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