Question

If $l$ is the length of the median from the vertex A to the side BC of a $\Delta ABC,$ then
(a) $4{{l}^{2}}=2{{b}^{2}}+2{{c}^{2}}-{{a}^{2}}$
(b) $4{{l}^{2}}={{b}^{2}}+{{c}^{2}}+2bc\cos A$
(c) $4{{l}^{2}}={{a}^{2}}+4bc\cos A$
(d) $4{{l}^{2}}={{\left( 2s-a \right)}^{2}}-4bc{{\sin }^{2}}\left( \dfrac{A}{2} \right)$

Answer 1

Hint: First find the midpoint of side opposite to the vertex A i.e. side BC. By definition of median find a relation between the sides which includes the length of median. By solving that condition, you can use cosine rule in between. By that you get a relation between all lengths of sides and length of median. By option’s just change the terms algebraically to match with any of the cosine rule:

Complete step-by-step answer:

${{b}^{2}}+{{c}^{2}}-{{a}^{2}}=2bc\operatorname{cosA}$

Median: The line joining the vertex with the midpoint of the opposite side is called median.
Altitude: The perpendicular line dropped from the vertex on to the opposite side of the vertex is called altitude

Let the midpoint of BC be point D.

Let the foot of perpendicular from A on to BC be E. From above we can draw triangle ABC as

By basic knowledge of algebra, we can say Pythagoras theorem:

Triangle ABC right angled at B: $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$

We can see in the figure that triangle AEB, AEC are right angle triangles using Pythagoras theorem here, we get:

$A{{B}^{2}}+A{{C}^{2}}=B{{E}^{2}}+A{{E}^{2}}+E{{C}^{2}}+A{{E}^{2}}$

By using the diagram, we can write the following terms:

BE = BD – ED; EC = ED + DC

By substituting these into the equation, we get that:

$A{{B}^{2}}+A{{C}^{2}}=2A{{E}^{2}}+{{\left( BD-ED \right)}^{2}}+{{\left( ED+DC \right)}^{2}}$

As BD = DC, we can apply following formula:

$\begin{align}

  & {{\left( a-b \right)}^{2}}+{{\left( a+b \right)}^{2}}=2\left( {{a}^{2}}+{{b}^{2}} \right) \\

 & A{{B}^{2}}+A{{C}^{2}}=2\left( A{{E}^{2}}+A{{D}^{2}}+B{{D}^{2}} \right) \\

 & \Rightarrow A{{B}^{2}}+A{{C}^{2}}=2\left( A{{D}^{2}}+B{{D}^{2}} \right) \\

\end{align}$

In our question we have sides. So, by substituting we get:

${{c}^{2}}+{{b}^{2}}=2\left[ {{l}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}} \right]\Rightarrow 2{{c}^{2}}+2{{b}^{2}}=4{{l}^{2}}+{{a}^{2}}$

By subtracting \[2{{a}^{2}}\] on both sides we get,

\[2{{c}^{2}}+2{{b}^{2}}-2{{a}^{2}}=4{{l}^{2}}-{{a}^{2}}\]

By basic knowledge of triangles, we can say cosine rule:

${{b}^{2}}+{{c}^{2}}-{{a}^{2}}=2bc\cos A$

By substituting this into the equation, we convert it into:

$4bc\cos A=4{{l}^{2}}-{{a}^{2}}$

By adding ${{a}^{2}}$ on both sides of equation, we get

$4bc\cos A+{{a}^{2}}=4{{l}^{2}}$

Option (c) is correct.


Note: Be careful while using cosine rule as you must observe it is multiplied by 2, don’t forget to take 4 as coefficient of cosine.