AD is the median of triangle ABC and G divides AD in the ratio 2:1 .Prove that area (triangle AGB )= area (triangle BGC )=area (triangle AGC )=1÷3area (triangle ABC )
I did this for you on Sunday
http://www.jiskha.com/display.cgi?id=1451271437
and I assume that 1÷3 meant 1/3
To prove that the areas of triangles AGB, BGC, and AGC are equal to one-third of the area of triangle ABC, we can use the fact that the medians of a triangle divide it into six equal area triangles.
Here's a step-by-step explanation of how to prove this:
1. Recall that the medians of a triangle are the line segments connecting each vertex to the midpoint of the opposite side. In this case, AD is the median of triangle ABC, and G divides AD in the ratio 2:1.
2. Let M be the midpoint of BC (side opposite to vertex A). Since AD is the median, it divides BC in the ratio 1:1.
3. Now, let's consider triangle AGB. Since G divides AD in the ratio 2:1, we can consider AG as 2 parts and GD as 1 part.
4. According to the property of medians, triangle AGB is divided into four triangles of equal areas. Let's call these triangles GMA, GMB, GDC, and GDA.
5. The area of triangle AGB is equal to the sum of the areas of these four triangles (GMA + GMB + GDC + GDA).
6. Since GD is equal to 1 part and GM is equal to 1 part (as AD divides BC in the ratio 1:1), triangle GDC is equal in area to triangle GMA.
7. Similarly, triangle GDA is equal in area to triangle GMB, as GD is equal to 1 part and GM is equal to 1 part.
8. Therefore, the area of triangle AGB is 2 times the area of either GMA or GMB.
9. Using the same logic, we can show that the areas of triangles BGC and AGC are also 2 times the area of either GMB or GMA.
10. Since we have shown that the areas of triangles AGB, BGC, and AGC are each 2 times the area of either GMA or GMB, their areas are equal to each other.
11. Hence, the area of triangle AGB is equal to the area of triangle BGC, which is also equal to the area of triangle AGC.
12. Furthermore, since the sum of the areas of these three triangles is equal to the sum of the areas of GMA and GMB (each equal to one-third of the area of triangle ABC), we can conclude that the area of each triangle (AGB, BGC, and AGC) is equal to one-third of the area of triangle ABC.
Therefore, we have proven that the area of triangle AGB is equal to the area of triangle BGC, which is also equal to the area of triangle AGC, and each of these areas is equal to one-third of the area of triangle ABC.