Question

Prove that \[A{B^2} + A{C^2} = 2\left( {A{D^2} + B{D^2}} \right)\]

Answer 1

Hint:
Here we will first use Pythagoras theorem on the three right-angle triangles formed and add the two whose left hand side we want to form an equation. Next, we will use the third right-angle triangle to replace the terms on the right hand side as per our requirement. Finally, we will solve the equation with all the relations we have to prove the given equation.

Formula Used:
Pythagoras Theorem: \[{\left( B \right)^2} + {\left( P \right)^2} = {\left( H \right)^2}\], where, \[B\] is the base of the triangle, \[P\] is the perpendicular of the triangle and \[H\] is the hypotenuse of the triangle.

Complete step by step solution:
The triangle given to us is divided in three parts or three triangles. So, we will use Pythagoras theorem in each right-angle triangle. Now applying Pythagoras Theorem in \[\Delta ABE\], we get
\[A{B^2} = A{E^2} + B{E^2}\]……..\[\left( 1 \right)\]
Now applying Pythagoras Theorem in \[\Delta ACE\], we get
\[A{C^2} = A{E^2} + C{E^2}\] ……..\[\left( 2 \right)\]
Now applying Pythagoras Theorem in \[\Delta AED\], we get
\[A{D^2} = A{E^2} + E{D^2}\]
\[A{E^2} = A{D^2} - E{D^2}\]……\[\left( 3 \right)\]
Now adding equation \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[\begin{array}{l}A{B^2} + A{C^2} = A{E^2} + B{E^2} + A{E^2} + C{E^2}\\ \Rightarrow A{B^2} + A{C^2} = 2A{E^2} + B{E^2} + C{E^2}\end{array}\]
Now, we will convert \[AE\] into \[AD\]in above equation from equation \[\left( 3 \right)\], we get
\[ \Rightarrow A{B^2} + A{C^2} = 2\left( {A{D^2} - E{D^2}} \right) + B{E^2} + C{E^2}\]…..\[\left( 4 \right)\]
Now as we can see from the diagram that
\[\begin{array}{l}BE = BD - ED\\CE = CD + ED\end{array}\]
Substituting above values in equation \[\left( 4 \right)\], we get
\[\begin{array}{l} \Rightarrow A{B^2} + A{C^2} = 2\left( {A{D^2} - E{D^2}} \right) + {\left( {BD - ED} \right)^2} + {\left( {CD + ED} \right)^2}\\ \Rightarrow A{B^2} + A{C^2} = 2A{D^2} - 2E{D^2} + B{D^2} + E{D^2} - 2BDED + C{D^2} + E{D^2} + 2CDED\end{array}\]
On solving further we get,
\[ \Rightarrow A{B^2} + A{C^2} = 2A{D^2} + B{D^2} + C{D^2} + 2ED\left( {CD - BD} \right)\]…..\[\left( 5 \right)\]
As, \[AD\] is the median of \[BC\], therefore \[CD = BD\].
Substituting above value in equation \[\left( 5 \right)\], we get
\[ \Rightarrow A{B^2} + A{C^2} = 2A{D^2} + B{D^2} + B{D^2} + 2ED\left( {BD - BD} \right)\]
Adding the like terms, we get
\[ \Rightarrow A{B^2} + A{C^2} = 2A{D^2} + 2B{D^2}\]
Taking 2 common from both the sides, we get

\[ \Rightarrow A{B^2} + A{C^2} = 2\left( {A{D^2} + B{D^2}} \right)\]
Hence Proved

Note:
Here, we have solved the question using Pythagoras theorem. Pythagoras theorem states that the square of the hypotenuse is equal to the sum of squares of the other two sides of a right angled triangle. This theorem can only be applied to right angled theorems. It is generally used to find the length of an unknown side and angle of a triangle. The side opposite to the right angle is known as Hypotenuse and it is the longest side.