Question

BL and CM are medians of triangle ABC right angled at A. Prove that 4(BL^2+CM^2)= = 5BC^2

Answer 1

let,AM=MB=x and AL=LC=y.
now,LHS=4(BL^2+CM^2)
=4[{(2x)^2+(y)^2}+{(2y)^2+(x)^2}] (by pythagoras theorem)
=4[4x^2+y^2+4y^2+x^2]
=4[5x^2+5y^2]
=20x^2+20y^2 …… (1)
and now,RHS=5BC^2
=5[(2x)^2+(2y)^2] (since by pythagoras theorem h^2=b^2+p^2,p=perpndicular, b=base and h=hypotenuse)
=5[4x^2+4y^2]
=20x^2+20y^2 ……(2)
LHS=RHS
hence,proved

Answer 2

Answer:

Step-by-step explanation:

Given : ABC right angle triangle at A where BL and CM are the medians

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)

In triangle ABC

BC^2 = AB^2 + AC^2

In triangle BAL

BL^2 =AB^2 +AL^2

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

On further little solving steps

4BL^2 =4AB^2 +AC^2

Then in triangle MAC

CM^2 =AM^2 +AC^2

Doing same process here

We get

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

please refer next part of answer in pic which is up made for you

Hope it helps you..............!!