Question

In a triangle \[ABC\], \[AD\] is drawn perpendicular to \[BC\]. How do I prove that \[A{B^2} - B{D^2} = A{C^2} - C{D^2}\]?

Answer 1

Hint: To solve this question, we will first draw a rough figure from the data given in the question. We will have two right triangles when we draw the perpendicular. In each right triangle we will apply the Pythagoras theorem. Then we will manipulate the data to prove the above relation.

Complete step by step answer:

In the above triangle ABC, AD is perpendicular to BC. We have two right triangles in the triangle ABC. Triangle ADB and triangle ADC are right angle triangles. So, we will apply the Pythagoras theorem in these two triangles.Pythagoras theorem states that in a right triangle;
\[{\left( {{\text{hypotenuse}}} \right)^2} = {\left( {{\text{base}}} \right)^2} + {\left( {{\text{perpendicular}}} \right)^2}\]

In triangle ADB,
Hypotenuse \[ = AB\]
So, applying the Pythagoras theorem, we get;
\[ \Rightarrow A{B^2} = A{D^2} + B{D^2}\]
On rearranging we get;
\[ \Rightarrow A{D^2} = A{B^2} - B{D^2}.....(1)\]

Similarly, in triangle ADC, we will apply the Pythagoras theorem. So, we have;
\[ \Rightarrow A{C^2} = A{D^2} + C{D^2}\]
On rearranging we get;
\[ \Rightarrow A{D^2} = A{C^2} - C{D^2}.....(2)\]
In equation \[(2)\] and \[(1)\] we can see that the LHS of both the equations are the same. So, their RHS must be the same. So, on equating the RHS we have;
\[ \Rightarrow A{B^2} - B{D^2} = A{C^2} - C{D^2}\]
Hence, proved.

Note: Suppose in any triangle AD is the bisector of the side BC. Also, if the sides AB and AC are equal then, then we can directly prove the above relation. There we will not need to use the Pythagoras theorem. Because the length of AB and AC will be equal and also since, AD is the bisector of the third side so, BD will be equal to CD. Then we will have;
\[A{B^2} = A{C^2}\] and \[B{D^2} = C{D^2}\].
On subtracting we will get;
\[ \therefore A{B^2} - B{D^2} = A{C^2} - C{D^2}\]