Question

In ∆ABC, ∠BAC = 90° , seg BL and seg CM are medians of ∆ABC. Then prove that 4 (BL² + CM² ) = 5 BC².

Answer 1

Given : ∠BAC = 90° 

seg BL and seg CM are the medians.

To prove: 4(BL² + CM² ) = 5BC²

In ∆BAL, ∠BAL 90° [Given] 

∴ BL² = AB² + AL² (i) [Pythagoras theorem] 

In ∆CAM, ∠CAM = 90° [Given] 

∴ CM² = AC² + AM² (ii) [Pythagoras theorem] 

∴ BL² + CM² = AB² + AC² + AL² + AM² (iii) [Adding (i) and (ii)] 

Now, AL = 1/2 AC and AM = 1/2 AB (iv) [seg BL and seg CM are the medians] 

∴ BL² + CM² 

= AB² + AC² + ( 1/2 AC)² + (1/2 AB)² [From (iii) and (iv)]

= AB² + AC² + AC²/4  + AB²/4 

= AB² + AB²/4 + AC² + AC²/4

= (5AB²⁾/4 + 5AC²/4

= BL² + CM² = 5/4(AB² + AC²)

∴ 4(BL² + CM² ) = 5(AB² + AC² ) (v) 

In ∆BAC, ∠BAC = 90° [Given] 

∴ BC² = AB² + AC² (vi) [Pythagoras theorem] 

∴ 4(BL² + CM² ) = 5BC² [From (v) and (vi)]