Question

एक चित्र में BL तथा CM समकोण त्रिभुज ABC की मध्यिकाए है तथा इस त्रिभुज का कोण A समकोण है सिद्ध कीजिए कि 4(BL)^2+4(CM)^2=5(BC)^2

Answer 1

$\textbf{Pythagors theorem:}$

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

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

$\textbf{Given:}$

$\text{BL and CM are medians of right triangle ABC}$

$\textbf{To prove:}$

$4(BL^2+CM^2)=5\,BC^2$

$\textbf{Solution:}$

$\text{Since BL and CM are medians of right triangle ABC,}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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