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

To prove: A circle drawn with $Q$ as centre, will pass through $A,B\text{and}O(i.e.QA=QB=QO)$

Proof:

As angle formed in the semicircle is a right angle.

$\Rightarrow \angle \mathrm{AOB}=90°$

Also the diagonal intersect at right angle.

$\Rightarrow \angle \mathrm{AOB}=\angle \mathrm{BOC}=\angle \mathrm{COD}=\angle \mathrm{DOA}=90°$

Thus, it is concluded that point $O$ lies on the circle.

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