Question

\[D\] and \[E\] are points on the sides \[CA\] and \[CB\] respectively of a \[\Delta ABC\] right angled at \[C\]. Prove that \[A{E^2} + B{D^2} = A{B^2} + D{E^2}\].

Answer 1

Hint: The given triangle is a right angle triangle, after joining the given points we will get three more right angle triangles.
So we can apply Pythagoras theorem for all the right angled triangles and we get four identities. Putting the values in the left hand side we can derive the right hand side.

Complete step-by-step answer:
It is given that a \[\Delta ABC\] is right angled at \[C\].
Also we have, \[D\] and \[E\] are points on the sides \[CA\] and \[CB\] respectively.

We need to prove that, \[A{E^2} + B{D^2} = A{B^2} + D{E^2}\].
Let us join the points \[D\] with \[E\] and \[B\] and the point \[E\] and \[A\].
Since \[\Delta ABC\] is a right angled triangle and \[\angle ACB = 90^\circ \] .
We can apply Pythagoras theorem for the right angled triangle ABC,
We get,
\[A{B^2} = A{C^2} + B{C^2} \ldots \ldots \left( 1 \right)\]
Since \[\Delta BCD\] is a right angled triangle and \[\angle DCB = 90^\circ \] .
We can apply Pythagoras theorem for the right angled triangle \[BCD\],
We get,
\[B{D^2} = D{C^2} + B{C^2} \ldots \ldots \left( 2 \right)\]
Since \[\Delta ACE\] is a right angled triangle and \[\angle ACE = 90^\circ \] .
We can apply Pythagoras theorem for the right angled triangle \[ACE\],
We get,
\[A{E^2} = A{C^2} + C{E^2} \ldots \ldots \left( 3 \right)\]
Since \[\Delta DCE\] is a right angled triangle and \[\angle DCE = 90^\circ \] .
We can apply Pythagoras theorem for the right angled triangle \[DCE\],
We get,
\[D{E^2} = D{C^2} + C{E^2} \ldots \ldots \left( 4 \right)\]
Putting the values from (3) and (2), we have,
L.H.S. =
\[ \Rightarrow A{E^2} + B{D^2}\]
Using the triangle diagram,
\[ \Rightarrow A{C^2} + C{E^2} + D{C^2} + B{C^2}\]
Rearranging the terms,
\[ \Rightarrow (A{C^2} + B{C^2}) + (C{E^2} + D{C^2})\]
Again using the triangle diagram,
\[ \Rightarrow A{B^2} + D{E^2}\]
= R.H.S.
Hence, \[A{E^2} + B{D^2} = A{B^2} + D{E^2}\].

Note: In this question we may go wrong on substituting sides of the triangle in to get the required result. So, be concentrated on that calculation when we use that method.
Pythagoras theorem states that,

If ABC is a right angled triangle then,
\[{{\text{(Hypotenuse)}}^{\text{2}}}{\text{ = (Height}}{{\text{)}}^{\text{2}}}{\text{ + (Base}}{{\text{)}}^{\text{2}}}\]
\[A{C^2} = A{B^2} + B{C^2}\]