In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that: 4(BL2 + CM2) = 5 BC2
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 11 | Page 45
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
Explanation:-
We know that in a right-angled triangle, the length of the median drawn to the hypotenuse is half the length of the hypotenuse. Therefore, we have:
BL = BC/2 and CM = AC/2
Squaring both sides, we get:
BL^2 = BC^2/4 and CM^2 = AC^2/4
Multiplying by 4, we get:
4BL^2 = BC^2 and 4CM^2 = AC^2
Adding these two equations, we get:
4BL^2 + 4CM^2 = BC^2 + AC^2
But we know that BC^2 + AC^2 = AB^2 (by the Pythagorean theorem).
Therefore, we get:
4BL^2 + 4CM^2 = AB^2
But we also know that BL and CM are medians of the triangle, so we have:
2BL^2 + 2CM^2 = AC^2 + BC^2/2
Adding these two equations, we get:
6BL^2 + 6CM^2 = AB^2 + AC^2 + BC^2/2
But we know that AB^2 + AC^2 + BC^2/2 = 5BC^2/4 (by the Pythagorean theorem).
Therefore, we get:
6BL^2 + 6CM^2 = 5BC^2/4
Dividing both sides by 3, we get:
2BL^2 + 2CM^2 = 5BC^2/12
Multiplying by 2, we get:
4BL^2 + 4CM^2 = 5BC^2/6
Hence, we have proved that:
4(BL^2 + CM^2) = 5BC^2
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 11 | Page 45