Question

BL and CM are medians on sides AC and AB of triangle ABC, right angled at A. prove that 4(BL^2 +CM^2)=5BC^2

Answer 1

Please refer to the attachment

Answer 2

Let's consider the triangle ABC, where A is the right angle, B and C are the remaining vertices, and AB and AC are the sides opposite to angles B and C, respectively. BL and CM are the medians on sides AC and AB, respectively.

Since BL and CM are medians, they bisect each other at the centroid G of the triangle. Let's denote by D and E the feet of the perpendiculars from G to AB and AC, respectively. Then, by the Pythagorean Theorem, we have:

$BD^2 + GD^2 = BL^2 and GE^2 + CE^2 = CM^2$

Adding these two equations, we get:

$BD^2 + GD^2 + GE^2 + CE^2 = BL^2 + CM^2$

Since triangle ABC is right-angled at A, we have:

$GE^2 + GD^2 = AB^2$

Also, $BD^2 + CE^2 = AC^2$

Substituting these two equations into the equation derived above, we get:

$AC^2 + AB^2 = BL^2 + CM^2$

Now, since triangle ABC is right-angled at A, we can use the Pythagorean Theorem to express$BC^2$ in terms of $AC^2$ and $AB^2$:

$BC^2 = AC^2 + AB^2$

Substituting this into the equation derived above, we get:

$BC^2 = BL^2 + CM^2$

Finally, multiplying both sides of this equation by 4, we get:

$4(BL^2 + CM^2) = 4 * BC^2 = 5 * BC^2$

This proves that $4(BL^2 + CM^2) = 5BC^2.$

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