Q5 of 35 Page 186

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

Let ABCD be a rhombus in which diagonals intersect at point O and a circle is drawn is drawn taking side CD as diameter

Now,


∠COD = 900


(Since, diagonals of rhombus intersect at 900)


Hence,


∠AOB = ∠BOC = ∠COD = ∠DOA = 900


Hence, point O has to lie on the circle


More from this chapter

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The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

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Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

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ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

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AC and BD are chords of a circle which bisect each other. Prove that:

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