If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral. Show that it is a rectangle.
Explanation:
Given \(AC\) and \(BD\) are the diameters of the circle.
We know that, 'An angle inscribed in a semi-circle is a .'
Therefore, \(\angle BAD = \angle BCD = \angle ABC = \angle ADC = \)\(^\circ\) -------(1)
Now consider the triangles \(\bigtriangleup ABC\) and \(\bigtriangleup BAD\)
\(AC=\) []
\(AB=\) []
\(\angle ABC = \angle \) [Each equal to \(90^\circ\)]
Then \(\bigtriangleup ABC \cong \bigtriangleup BAD \) [by ].
This implies, \(AD = BC\). [By C.P.CT]
similalry, \(AB=DC\).
Therefore \(AB = BC = CD = DA\) --------(2)
Therefore, from (1) and (2), opposite sides are equal and each angle is \(90^\circ\). Then the shape formed is a rectangle.
\(ABCD\) is a rectangle.