Question

Prove that the sum of the squares of the lengths of the medians of a triangle is three fourths the sum of the squares of the lengths of the sides.

Answer 1

Answer:

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

Step-by-step explanation:

I've attached an image showing a triangle divided into it's medians.

Now, from the attached image, we see that AB, BC & CD are the lengths of sides of triangle while AD, BE & CF are lengths of the 3 medians of the triangle.

Now, to prove our question, we will use appolonius theorem which states that "the sum of the squares of any of the two sides of a triangle is equal to twice the square on half the third side including twice the square on the median that bisects the third side.

Applying this theorem to the image attached, we have the following;

|AB|² + |AC|² = 2[|AD|² + |BC/2|²]

|AB|² + |BC|² = 2[|BE|² + |AC/2|²]

|AC|² + |BC|² = 2[|CF|² + |AB/2|²]

Adding the 3 equations above gives us;

2|AB|² + 2|BC|² + 2|AC|² = 2|AD|² + |BC|²/2 + 2|BE|² + |AC|²/2 + 2|CF|² + |AB|²/2

Collecting like terms;

(2|AB|² - |AB|²/2) + (2|BC|² - |BC|²/2) + (2|AC|² - |AC|²/2) = 2|AD|² + 2|BE|² + 2|CF|²

Thus gives;

(3/2)[|AB|² + |BC|² + |AC|²] = 2[|AD|² + |BE|² + |CF|²]

Multiply both sides by 2 to give;

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]