Question

Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer 1

Hint: Use the information that diagonals of a rhombus bisect each other at right angles. Also, non-adjacent sides are parallel in case of rhombus.

Complete step-by-step answer:

Let us consider rhombus ABCD with the two diagonals as AC and BD. Consider O as the point of intersection of diagonals. We know that diagonals of a rhombus bisect each other at right angles. We need to prove that circle drawn on AB as diameter will pass through O. Draw lines MN and XY such that $MN||AD$ and $XY||AB$ such that M and N are mid-points of AB and CD. $
  AB = DC \\
   \Rightarrow \dfrac{1}{2}AB = \dfrac{1}{2}DC \\
   \Rightarrow AM = DN{\text{ }}[{\text{M and N are mid - points of AB}}{\text{ and CD}}] \\
$
Similarly, $
  AX = OM \\
   \Rightarrow AN = ON = NB \\
$
A circle drawn with N as centre and radius AN pass-through A, O and B. Obtained circle is the required circle.

Note: Using the basic properties of rhombus is the key for this question. Also, which property should be used. This thing comes by practice.