Question

In ∆ ABC angle BAC = 90° seg BL and seg CM are medians of ∆ABC. Then prove that: 4(BLsquare + CM square) = 5BC square

Answer 1

Answer:

BL and CM are medians

\text{Then,$AL=\frac{AC}{2}$ and $AM=\frac{AB}{2}$}Then,AL=2AC and AM=2AB .........(1)

\text{In $\triangle$BAL, by pythagoras theorem}In △BAL, by pythagoras theorem

BL^2=AB^2+AL^2BL2=AB2+AL2 .......(2)

\text{In $\triangle$MAC, by pythagoras theorem}In △MAC, by pythagoras theorem

CM^2=AM^2+AC^2CM2=AM2+AC2 .......(3)

\text{In $\triangle$BAC, by pythagoras theorem}In △BAC, by pythagoras theorem

BC^2=AB^2+AC^2BC2=AB2+AC2 .......(4)

\text{Adding (2) and (3), we get}Adding (2) and (3), we get

BL^2+CM^2=AB^2+AL^2+AM^2+AC^2BL2+CM2=AB2+AL2+AM2+AC2

BL^2+CM^2=(AB^2+AC^2)+AL^2+AM^2BL2+CM2=(AB2+AC2)+AL2+AM2

\text{Using (4), we get}Using (4), we get

BL^2+CM^2=BC^2+AL^2+AM^2BL2+CM2=BC2+AL2+AM2

\text{Using (1), we get}Using (1), we get

BL^2+CM^2=BC^2+(\frac{AC}{2})^2+(\frac{AB}{2})^2BL2+CM2=BC2+(2AC)2+(2AB)2

BL^2+CM^2=BC^2+\frac{AC^2}{4}+\frac{AB^2}{4}BL2+CM2=BC2+4AC2+4AB

BL2+CM2=1/4(AB2+AC2)

BL2+CM2=BC2+1/4BC2

BL2+CM2=4B2+BC2/4

BL2+CM2=5BC2/4

4(BL2+CM2)=5BC2

Answer 2

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