Question

In the figure, BL and CM are the medians of ABC, right angled at A. Prove that 4(BL2+ CM2) = 5 BC2. (refer the figure) guys please answer about this question I don't know how to solve this question so please answer this and those who will answer this in 5 minutes he or she will be counted by me as a brainliest ​

Answer 1

in tringle BLA by Pythagoras theorem

AB2 +AL2=BL2

butAL=1/2 AC

AB2+(1/2AC)2=BL2

AB2+1/4AC2=BL2

multipleying both of side by 4 we get

4AB2+AC 2=4BL2 ....(1)

now in tringle MAC by Pythagoras theorem

AC2 + AM2=CM2

but AM =1/2AB

AC2 +(1/2AB)=CM2

AC2 + 1/4 AB =CM2

multipleying both side by 4 we get

4AC2+AB2=4CM2.....(2)

adding 1 and 2

4AB2 +AC2=4BL2

4AC2 +AB2=4CM2

4(CM2+BL2)=AB2 + AC2+ 4AB2 +4AC2

but AB2 +AC2=BC2

therefore 4(BL2 +CM2)= BC2 + 4BC2

hence 4(BL2 +CM2)=5BC2

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