Question

$BL$ and $CM$ are the medians of a triangle $ABC$, right angled at $A$, then which one is true

• A. $4(B{L}^2+C{M}^2)=5B{C}^2$
• B. $4(B{L}^2+C{M}^2)=3B{C}^2$
• C. $2(B{L}^2+C{M}^2)=5B{C}^2$
• D. $2(B{L}^2+C{M}^2)=3B{C}^2$

Answer 1

The correct option is A $4(B{L}^2+C{M}^2)=5B{C}^2$

Given: $\Delta ABC$ right angled at $A$, i.e. $\angle A+{90}^0$ where $BL$ and $CM$ are the medians

Proof:- Since $BL$ is median,

$AL=CL=\frac{1}{2}AC$...........(1)

Similarly, $CM$ is the median

$AM=MB=\frac{1}{2}AB$...........(2)

We know that, by Pythagoras theorem

$(hypotimise{)}^2=(height{)}^2+(Base{)}^2$

In $\Delta BAC$,

$(BC{)}^2=(AB{)}^2+\left(AC\right)$.........(1)

In $\Delta BAL$,

$(BL{)}^2=(AB{)}^2+\left(AL\right)$.........(1) $From\left(1\right):AL=\frac{1}{2}AC$

$=(AB{)}^2+{\left(\frac{AC}{2}\right)}^2$

$=(AB{)}^2+\frac{(AC{)}^2}{4}{)}^2$

$(BL{)}^2=\frac{4(AB{)}^2+(AC{)}^2}{4}$

$4B{L}^2=4(AB{)}^2+A{C}^2$,,,,,,,,,(ii)

In $\Delta MAC$

$(CM{)}^2=(AM{)}^2+(AC{)}^2$ $From\left(2\right)AM=\frac{1}{2}AB$

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

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

$C{M}^2=\frac{(AB{)}^2+4(AC{)}^2}{4}$

$4(CM{)}^2=(AB{)}^2+4(AC{)}^2$............(iii)

adding (ii) and (iii)

$4(BL{)}^2+4(CM{)}^2=4(AB{)}^2+(AC{)}^2+(AB{)}^2+4(AC{)}^2$

$4(BL{)}^2+(CM{)}^2=5(AB{)}^2+5(AC{)}^2$

$4(BL{)}^2+(CM{)}^2=5(AB{)}^2+5(AC{)}^2$

$4(BL{)}^2+(CM{)}^2=5(BC{)}^2$

$From(BC{)}^2=(AB{)}^2+(AC{)}^2$

Hence, proved