BL& CM are medians of a triangle ABC 90 degree at A .Prove that4(BL2+C

1.	BL& CM are medians of a triangle ABC 90 degree at A .Prove that4(BL2+CM2)=5BC2.
Answer» BL and CM are medians of a {tex}\\triangle {/tex} ABC in which {tex}\\angle{/tex} A=900From {tex}\\triangle {/tex}ABC, BC2 = AB2 + AC2 ....(i)From right angled {tex}\\vartriangle {/tex} ABL,BL2 = AL2 + AB2i.e {tex}B{L^2} = {\\left( {\\frac{{AC}}{2}} \\right)^2} + A{B^2}{/tex}{tex}\\Rightarrow {/tex} 4BL2= AC2 + 4AB2 .....(ii)From right-angled {tex}\\triangle {/tex}CMA,CM2 = AC2 + AM2i.e {tex}C{M^2} = A{C^2} + {\\left( {\\frac{{AB}}{2}} \\right)^2}{/tex}[mid-point]{tex}\\Rightarrow {/tex}{tex}C{M^2} = A{C^2} + \\frac{{A{B^2}}}{4}{/tex}{tex}\\Rightarrow {/tex}{tex}4C{M^2} = 4A{C^2} + A{B^2}{/tex} .....(iii)Adding (ii) and (iii), we geti.e.4(BL2 + CM2) = 5(AC2 + AB2) = 5BC2 [From (i)]

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