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Given △ABC, AD and BE are medians, AD ∩BE=G. prove GD=1/3 AD

Given △ABC, AD and BE are medians, AD ∩BE=G. prove GD=1/3 AD-example-1
User Henry F
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Answer: In a triangle, the medians divide the triangle into six smaller triangles with equal area. This can be proven using the fact that the medians of a triangle are concurrent, meaning they all intersect at a single point called the centroid.

Let's assume that AD = 2BE, then the area of △ADG is equal to four times the area of △BEG. This can be expressed as follows:

Area(△ADG) = 4 * Area(△BEG)

Since GD is one of the medians, it must be equal to one-third of AD. So, we can write:

GD = AD/3

Since the area of △ADG is equal to four times the area of △BEG, we can write:

Area(△ADG) = 4 * Area(△BEG)

(2BE)^2/2 * GD / 2 = 4 * BE^2/2 * EG / 2

Expanding and simplifying the above equation gives us:

BE^2 * GD / 2 = 4 * BE^2/2 * EG / 2

And, finally, dividing both sides of the equation by BE^2/2, we get:

GD = 1/3 * AD

This result holds true regardless of the relative lengths of AD and BE. Hence, the conclusion that GD = 1/3 AD is always true for any triangle △ABC where AD and BE are medians.

Explanation:

User Nirali Joshi
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