Question

In a quadrilateral ABCD, given that \[\angle A + \angle D = 90^\circ\]. Prove that \[A{C^2} + B{D^2} = A{D^2} + B{C^2}\].

Answer 1

Hint: In this problem, first draw a quadrilateral ABCD. The side AB and DC are produced to meet a point O. Now use Pythagoras theorem to prove the given expression.
Draw a quadrilateral ABCD and produce the sides AB and DC to meet at point O as shown below.

Complete step-by-step solution -

The sum of interior angles of a triangle is \[180^\circ\]. Therefore,
In \[\Delta AOD\],
\[ \angle A + \angle O + \angle D = 180^\circ \\
   \Rightarrow \angle O = 180^\circ - \left( {\angle A + \angle D} \right) \\
   \Rightarrow \angle O = 180^\circ - 90^\circ \\
   \Rightarrow \angle O = 90^\circ \\ \]
The side \[AD\] of the triangle \[\Delta AOD\] represents the hypotenuse of the triangle. Similarly, sides \[BC,AC\] and \[BD\] represent the hypotenuse of the triangles \[\Delta BOC\], \[\Delta AOC\] and \[\Delta BOD\] respectively.
Now, using Pythagoras theorem in \[\Delta AOD\], \[\Delta BOC\] , \[\Delta AOC\] and \[\Delta BOD\] we get,
\[\begin{gathered}
  A{D^2} = A{O^2} + O{D^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\
  B{C^2} = B{O^2} + O{C^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\
  A{C^2} = A{O^2} + O{C^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right) \\
  B{D^2} = O{D^2} + B{O^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 4 \right) \\
\end{gathered}\]

Now, add equation (1) and equation (2) as shown below.
\[ A{D^2} + B{C^2} = A{O^2} + O{D^2} + B{O^2} + O{C^2} \\
  A{D^2} + B{C^2} = \left( {A{O^2} + O{C^2}} \right) + \left( {B{O^2} + O{D^2}} \right) \\
  A{D^2} + B{C^2} = A{C^2} + B{D^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From equation (3) and (4)}}} \right) \\ \]
Hence proved.

Note: The key concept of these types of problems is the angle sum property of a triangle. Join the diagonals of the quadrilateral. Also, produce the sides AB and DC to meet at point O and then apply the Pythagoras theorem. Except adjacent sides of angle o all sides are to be hypotenuse for triangle AOD.