Question

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

Answer 1

Let ABCD be a rhombus whose diagonals AC and BD intersect at O.

We know that, diagonals of a rhombus intersect each other at right angle.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, circles with AB, BC, CD and DA as their respective diameters pass through O.  (Angle in a semi-circle is 90°)

Thus, the circles described on the four sides of a rhombus as their respective diameters, pass through the point of intersection of its diagonals.
Hence, proved.

Answer 2

Step-by-step explanation:

Sol: Consider ABCD be a rhombus whose diagonals AC and BD intersect at O.

Since diagonals of a rhombus intersect each other at right angle.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, AB, BC, CD and DA as diameters of circles passes through O. Angle in a semi-circle is 90°.

Hence the circles described on the four sides of a rhombus as diameter, passes through the point of intersection of its diagonals.