Question

In \[\Delta ABC\], if AD is the median, then show that \[A{B^2} + A{C^2} = 2\left( {A{D^2} + B{D^2}} \right)\]

Answer 1

Hint: First of all, carefully observe the given figure. Then we can notice that there are three right angled triangles. Use Pythagoras theorem to those three triangles and we get three equations. By suitable substitutions we can reach the solution of the given problem.

Complete step by step solution:
Given that \[AD\] is the median of \[\Delta ABC\].
So, clearly \[BD = CD...................................................\left( 1 \right)\]
Also, \[AE \bot BC\]
Now we have to prove that \[A{B^2} + A{C^2} = 2\left( {A{D^2} + B{D^2}} \right)\]
By Pythagoras theorem we know that, \[{\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}\]
In the Right \[\Delta ABE\], by applying the Pythagoras theorem, we have
\[A{B^2} = A{E^2} + B{E^2}.................................................\left( 2 \right)\]
And in Right \[\Delta ACE\], by applying Pythagoras theorem, we have
\[A{C^2} = A{E^2} + C{E^2}..............................................\left( 3 \right)\]
Adding \[\left( 2 \right)\] and \[\left( 3 \right)\], we get
\[
   \Rightarrow A{B^2} + A{C^2} = A{E^2} + B{E^2} + A{E^2} + C{E^2} \\
   \Rightarrow A{B^2} + A{C^2} = 2A{E^2} + B{E^2} + C{E^2}......................................................\left( 4 \right) \\
\]
In Right \[\Delta AED\], by applying Pythagoras theorem, we have
\[
  A{D^2} = A{E^2} + E{D^2} \\
  A{E^2} = A{D^2} - E{D^2}...................................................\left( 5 \right) \\
\]
Substituting equation \[\left( 5 \right)\] in equation \[\left( 4 \right)\], we get
\[
   \Rightarrow A{B^2} + A{C^2} = 2\left( {A{D^2} - E{D^2}} \right) + B{E^2} + C{E^2} \\
   \Rightarrow A{B^2} + A{C^2} = 2A{D^2} - 2E{D^2} + B{E^2} + C{E^2} \\
\]
Now, writing \[BE = BD - ED\] and \[CE = CD + ED\] from the figure, we have
\[
   \Rightarrow A{B^2} + A{C^2} = 2A{D^2} - 2E{D^2} + {\left( {BD - ED} \right)^2} + {\left( {CD + ED} \right)^2} \\
   \Rightarrow A{B^2} + A{C^2} = 2A{D^2} - 2E{D^2} + B{D^2} - 2\left( {BD} \right)\left( {ED} \right) + E{D^2} + C{D^2} + 2\left( {CD} \right)\left( {ED} \right) + E{D^2} \\
\]
Grouping the common terms, we have
\[
   \Rightarrow A{B^2} + A{C^2} = 2A{D^2} + B{D^2} + C{D^2} + 2E{D^2} - 2E{D^2} + 2ED\left( {CD - BD} \right) \\
   \Rightarrow A{B^2} + A{C^2} = 2A{D^2} + B{D^2} + C{D^2} + 2ED\left( {CD - BD} \right) \\
\]
But, from equation \[\left( 1 \right)\] i.e., \[CD = BD\], we get
\[
   \Rightarrow A{B^2} + A{C^2} = 2A{D^2} + B{D^2} + B{D^2} + 2ED\left( {BD - BD} \right) \\
  \therefore A{B^2} + A{C^2} = 2A{D^2} + 2B{D^2} \\
\]
Hence, proved.

Note: Median of a triangle bisects its base into two equal parts in length. Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides” i.e., \[{\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}\].