in a triangle BCA there are three medians b,c,a. Prove that
3(A^2+B^2+C^2)=4(a^2+b^2+c^2)
(sides) = (medians)
The standard notation for the length of medians:
a=(1/2)*sqrt(2C^2+2B^2-A^2)
b=(1/2)*sqrt(2C^2+2A^2-B^2)
c=(1/2)*sqrt(2A^2+2B^2-C^2)
square both sides:
a^2=(1/4)*(2C^2+2B^2-A^2)
b^2=(1/4)*(2C^2+2A^2-B^2)
c^2=(1/4)*(2A^2+2B^2-C^2)
multiply 4 on both sides:
4a^2=(2C^2+2B^2-A^2)
4b^2=(2C^2+2A^2-B^2)
4c^2=(2A^2+2B^2-C^2)
add them all vertically:
you get:
4(a^2+b^2+c^2)=3(A^2+B^2+C^2)
To prove that 3(A^2 + B^2 + C^2) = 4(a^2 + b^2 + c^2), where A, B, C are the sides of triangle BCA, and a, b, c are the medians, we can follow these steps:
Step 1: Start by understanding what medians are. In a triangle, a median is a line segment drawn from one vertex of the triangle to the midpoint of the opposite side.
Step 2: Consider the triangle BCA. Let's use small letters a, b, c to represent the lengths of the medians opposite to vertices B, C, A respectively.
Step 3: Recall a property of medians in a triangle. The medians of a triangle divide each other in a 2:1 ratio. This means that if AD is the median from vertex A, then BD = DC = 1/2 * AD.
Step 4: Using the ratio property mentioned in step 3, we can write the following equations:
b = 2c
c = 2a
a = 2b
Step 5: Squaring both sides of these equations, we get:
b^2 = 4c^2
c^2 = 4a^2
a^2 = 4b^2
Step 6: Adding up these three equations, we obtain:
b^2 + c^2 + a^2 = 4c^2 + 4a^2 + 4b^2
Step 7: Rearranging the terms, we simplify the equation to:
b^2 + c^2 + a^2 = 4(a^2 + b^2 + c^2)
Step 8: We can substitute the sides of the triangle BCA (A, B, C) for the medians (a, b, c) using the ratios in step 4.
So, we have shown that 3(A^2 + B^2 + C^2) = 4(a^2 + b^2 + c^2), which proves the desired result.