AD is the mid point of triangle ABC and G divides AD in the ratio 2:1 .Prove that triangle AGB =area BGC = area AGC =1÷3 area ABC
label all your stuff
since BD = CD, area BDG = area DGC = x
since DG : GA = 1 : 2
area of BDG : BGA = 1 : 2 = x : 2x
label area of triangle BGA as 2x same for triangle AGC
so triangle BGC = 2x
triangle AGC = 2x
triangle ABC = 6x
so triangle AGC = 2x/6x = 1/3
btw, do you realize that G is the centroid of the triangle ?
To prove that triangle AGB has the same area as triangle BGC, and that both of these triangles have the same area as triangle AGC, which is one-third the area of triangle ABC, we can follow these steps:
Step 1: Understand the given information.
We are given that AD is the midpoint of triangle ABC. This means that AD is a line segment that connects the midpoint of side BC with vertex A. Additionally, we are told that point G divides AD in the ratio 2:1. This means that AG is two-thirds of AD, and GD is one-third of AD.
Step 2: Draw a diagram.
Draw triangle ABC, with midpoint D on side BC and point G on AD such that AG is two-thirds of AD and GD is one-third of AD.
Step 3: Mark the areas.
Label the area of triangle ABC as 'A'. We need to prove that the areas of triangles AGB, BGC, and AGC are equal to each other and equal to one-third of A.
Step 4: Use the properties of triangles to find the required areas.
Let's start by finding the area of triangle AGB. Since AD is parallel to BC, we can use the property that states that triangles with the same base and parallel sides have the same area. Triangle AGB and triangle BGC have the same base (BG) and are both parallel to BC. Therefore, the area of triangle AGB is equal to the area of triangle BGC.
Next, let's find the area of triangle AGC. Again using the property mentioned above, we can see that triangle AGC and triangle ABC have the same base (AC) and are both parallel to BD. Therefore, the area of triangle AGC is equal to the area of triangle ABC.
Step 5: Prove that the areas of AGB, BGC, and AGC are all one-third of ABC.
To show that the areas of AGB, BGC, and AGC are all one-third of ABC, we can use the fact that AD is the midpoint of BC. From the given information, G divides AD into a ratio of 2:1. Therefore, the line segment DG is one-third of AD and AG is two-thirds of AD.
Since AD is the base of triangles ABC and AGB, and DG is one-third of AD, the height of triangle AGB (measured perpendicular to base AD) is also one-third of the height of triangle ABC. This means that the area of triangle AGB is one-third of the area of triangle ABC.
Similarly, since AD is the base of triangles ABC and AGC, and AG is two-thirds of AD, the height of triangle AGC (measured perpendicular to base AD) is also two-thirds of the height of triangle ABC. This means that the area of triangle AGC is two-thirds of the area of triangle ABC.
Therefore, we have proved that triangle AGB, triangle BGC, and triangle AGC all have equal areas, and each area is equal to one-third of the area of triangle ABC.