#### Question

In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:

4(BL^{2 }+ CM^{2}) = 5 BC^{2}

^{}

#### Solution

**Given : **Δ ABC right angled at A i.e; A = 90°. where BL and CM are the median.

**To Prove : **4( BL^{2} + CM^{2 }) = 5BC^{2}

**Proof : **Since BL is the median,

AL = CL = `1/2` AC ...(1)

Similarly, CM is the median

AM = MB = `1/2`AB ...(2)

We know that, by pythagoras theorem

(Hypotenuse)^{2} = (Height)^{2} + (Base)^{2}

In ΔBAC,**(BC) ^{2} = (AB)^{2} + (AC)^{2 } ...(4)**

In ΔBAL,

(BL)^{2} = AB^{2} + AL^{2 } ...(From 1)

`("BL")^2 = AB^2 + ((AC)/2)^2`

`("BL")^2 = AB^2 + (AC)^2/4`

`("BL")^2 = [ 4("AB")^2 + ("AC")^2 ]/4`

**`4"BL"^2 = 4("AB")^2 + "AC"^2` ...(5)**

In ΔMAC,

`("CM")^2 = ("AM")^2 + ("AC")^2` ...( From 2)

`("CM")^2 = (("AB")/2)^2 + ("AC")^2`

`("CM")^2 = [("AB")^2 + 4("AC")^2 ]/4`

**`4"CM"^2 = "AB"^2 + 4"AC"^2` ...(6)**

From 4,5 and 6

(BC)^{2} = (AB)^{2} + (AC)^{2 }

`4"BL"^2 = 4("AB")^2 + "AC"^2`

`4"CM"^2 = "AB"^2 + 4"AC"^2`

Adding 5 and 6,

`4"BL"^2 + 4"CM"^2 = 4"AB"^2 + "AC"^2 + "AB"^2 + 4"AC"^2`

`4( "BL"^2 + "CM"^2 ) = 5"AB"^2 + 5"AC"^2`

`4( "BL"^2 + "CM"^2 ) = 5( "AB"^2 + "AC"^2 )`

`4( "BL"^2 + "CM"^2 ) = 5"BC"^2`