Question.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Solution:
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
$\therefore \angle C O D=90^{\circ}$
Also, in rhombus, the diagonals intersect each other at $90^{\circ}$.
$\angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}$
Clearly, point $O$ has to lie on the circle.
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
$\therefore \angle C O D=90^{\circ}$
Also, in rhombus, the diagonals intersect each other at $90^{\circ}$.
$\angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}$
Clearly, point $O$ has to lie on the circle.