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: Construction: Draw \[AE \bot BC\]
Pythagoras theorem: In a right angle triangle, the square of a hypotenuse side is equal to the sum of the square of the other two sides.
That is, in \[\Delta ABC,\,\,A{C^2} = A{B^2} + B{C^2}\]

Formula used:
Algebraic identities:-
\[
  {(a + b)^2}\, = \,{a^{2\,}}\, + \,2ab\, + \,{b^2}\,\,\,{\text{and}} \\
  {(a - b)^2} = \,{a^2}\, - \,2ab + \,{b^2} \\
 \]

Complete step-by-step solution:
To Prove \[A{B^2} + A{C^2} = 2\left( {A{D^2} + B{D^2}} \right)\]

From the above, figure, in the right triangle\[AED\,{\text{and}}\,AEC\], using Pythagoras theorem,
We write the expressions as follows,
\[A{B^2} + A{C^2} = B{E^2} + A{E^2} + E{C^2} + A{E^2}\]
Clearly, it is seen that from above expression, that we can add the like terms and that is
\[A{B^2} + A{C^2} = B{E^2} + 2A{E^2} + E{C^2}\]
Now, we have to simplify
\[BE\] As \[BE\, = \,BD\, - \,ED\]
And also, we simplify \[EC\] and it can be written as follows,
\[EC\, = \,ED\, + \,DC\]
Now, we can arrive at an equation as
\[A{B^2}\, + \,A{C^2}\, = \,2\,A{E^2}\, + \,{\left( {BD\, - \,ED} \right)^2}\, + \,{\left( {ED\, + \,DC} \right)^2}\]
We know that the algebraic identities:-
\[
  {(a + b)^2}\, = \,{a^{2\,}}\, + \,2ab\, + \,{b^2}\,\,\,{\text{and}} \\
  {(a - b)^2} = \,{a^2}\, - \,2ab + \,{b^2} \\
 \]
So, we can expand the above equation using the algebraic identities as follows:-
\[A{B^2}\, + A{C^2}\, = \,2A{E^2}\, + \,B{D^2}\, + \,E{D^2}\, - \,2(BD.ED)\, + \,E{D^2}\,D{C^2} + \,2\,(ED.DC)\]
Since, given that \[AD\] is median of the \[\Delta ABC\]
Therefore, \[BD\, = \,DC.\]
Also, \[BD.ED\, = \,ED.DC = \,0\] (By perpendicular Condition)
From the above Expression, we can re write the equation as,
\[A{B^{2\,}}\, + \,A{C^2}\, = \,2A{E^2}\,D{C^2}\, + \,2E{D^2}\, + \,D{C^2}\,\,\,\,\left( {\therefore \,BD = DC} \right)\]
Now, adding the two like terms as.
\[A{B^{2\,}}\, + \,A{C^2}\, = \,2A{E^2} + \,2E{D^2}\, + \,2D{C^2}\]
By Pythagoras theorem, we have in \[\Delta AED,\]
\[A{D^2}\, = \,A{E^2}\, + \,E{D^2}\]
Thus, we now have the equation by taking \[2\]as common out,
\[A{B^{2\,}} + \,A{C^2}\, = \,2\,\left( {A{E^2}\, + \,E{D^2}\,B{D^2}} \right)\]
Thus, by applying the Pythagoras theorem in the above which we have stated will become the equation as,
\[A{B^{2\,}} + \,A{C^2}\, = \,2\,\left( {A{D^2} + \,B{D^2}} \right)\]
Hence it is proved.

Note: In the above proof, we stated the condition as perpendicular, in the way that \[BD\,.\,ED\, = \,0\] and \[ED\,.\,DC\, = \,0\] is because as we constructed that \[AE \bot BC\]from the diagram, so that\[\angle E\, = \,{90^ \circ }\].
Thus, we have to keep an eye on the algebraic identities that help out the other equation hidden and give results accordingly as required. Also the perpendicular condition is helpful to use because whenever an angle is \[{90^ \circ }\] in any of the side angles then we say that it is perpendicular.