Question

Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.

Answer 1

Given: ABCD is a rhombus. Four circles are drawn on the sides AB, BC, CD and DA respectively.

To prove: The circles pass through the point of intersection of the diagonals of the rhombus ABCD.

Proof: ABCD is a rhombus whose diagonals AC and BD intersect each other at O.

$\because$ The diagonals of a rhombus bisect each other at right angles.

$\therefore \angle AOB=\angle BOC=\angle COD=\angle DOA={90}^o$

Now, as $\angle AOB={90}^o$, a circle described on AB as diameter will pass through O.

Similarly, the circles on BC, CD and DA as diameter, will also pass through O.