Question

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

Answer 1

Here, ABCD is a rhombus; we have to prove the four circles described on the four sides of any rhombusABCD pass through the point of intersection of its diagonals AC and BD.

Let the diagonals AC and BD intersect at O.

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

${\displaystyle \angle AOB=}$90°

${\displaystyle \angle BOC=}$90°

${\displaystyle \angle COD}$ = 90°

${\displaystyle \angle AOD}$ = 90° 

Now,  ${\displaystyle \angle AOB}$ = 90 means that circle described on AB as diameter passes through O.

Similarly the remaining three circles with BC, CD and AD as their diameter will also pass through O.

Hence, all the circles with described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.