Question

In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC²

Answer 1

$\textbf{Pythagors theorem:}$

$\text{In a right angled square on the hypotenuse is equal}$

$\text{to sum of the squares on the other two sides.}$

$\textbf{Given:}$

$\text{BL and CM are medians}$

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

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

$BL^2=AB^2+AL^2$.......(2)

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

$CM^2=AM^2+AC^2$.......(3)

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

$BC^2=AB^2+AC^2$.......(4)

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

$BL^2+CM^2=AB^2+AL^2+AM^2+AC^2$

$BL^2+CM^2=(AB^2+AC^2)+AL^2+AM^2$

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

$BL^2+CM^2=BC^2+AL^2+AM^2$

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

$BL^2+CM^2=BC^2+(\frac{AC}{2})^2+(\frac{AB}{2})^2$

$BL^2+CM^2=BC^2+\frac{AC^2}{4}+\frac{AB^2}{4}$

$BL^2+CM^2=BC^2+\frac{1}{4}[AB^2+AC^2]$

$BL^2+CM^2=BC^2+\frac{1}{4}BC^2$

$BL^2+CM^2=\frac{4\,BC^2+BC^2}{4}$

$BL^2+CM^2=\frac{5\,BC^2}{4}$

$\implies\boxed{\bf\,4(BL^2+CM^2)=5\,BC^2}$

Answer 2

Answer:

\textbf{Pythagors theorem:}Pythagors theorem:

\text{In a right angled square on the hypotenuse is equal}In a right angled square on the hypotenuse is equal

\text{to sum of the squares on the other two sides.}to sum of the squares on the other two sides.

\textbf{Given:}Given:

\text{BL and CM are medians}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+4AB2

BL^2+CM^2=BC^2+\frac{1}{4}[AB^2+AC^2]BL2+CM2=BC2+41[AB2+AC2]

BL^2+CM^2=BC^2+\frac{1}{4}BC^2BL2+CM2=BC2+41BC2

BL^2+CM^2=\frac{4\,BC^2+BC^2}{4}BL2+CM2=44BC2+BC2

BL^2+CM^2=\frac{5\,BC^2}{4}BL2+CM2=45BC2

\implies\boxed{\bf\,4(BL^2+CM^2)=5\,BC^2}⟹4(BL2+CM2)=5BC2