Question

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer 1

Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
∠BAD=\(\frac{1}{2}\)∠BOD=\(\frac{180^∘}{2}\)=90°
∠BCD+∠BAD=180∘ (Cyclic quadrilateral)  (Consider BD as a chord)
∠BCD=180∘−90∘=90°

∠ADC=\(\frac{1}{2}\)∠AOC=\(\frac{1}{2}\)(180)=90

∠ADC + ∠ABC = 180° (Cyclic quadrilateral) °+∠ABC = 180° 90°
∠ADC+∠ABC=180∘ (Opposite angles of a cyclic quadrilateral)
90 °+∠ABC=180∘
ABC = 90°       (Considering AC as a chord)
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.