Mathematics
In the figure (ii) given below, two equal chords AB and CD of a circle with center O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, prove that NOMP is a square.
Answer
In NOMP,
∠P = 90° (As chords intersect at right angles)
∠M = ∠N = 90° (Straight lines from center bisecting the chord are perpendicular to it.)
∠O = 360° - (∠M + ∠N + ∠P)
= 360° - (90° + 90° + 90°)
= 90°
Since, equal chords are equidistant from center,
∴ OM = ON.
Since, all angles = 90° and all sides are equal.
Hence, NOMP is a square.
Related Questions
In the figure (i) given below, C and D are centers of two intersecting circles. The line APQB is perpendicular to the line of centers CD. Prove that
(i) AP = QB
(ii) AQ = BP.
In the adjoining figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its center O, prove that AD bisects ∠BAC and ∠BDC.
In the figure (i) given below, two circles with centers C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.
In the figure (ii) given below, P is a point of intersection of two circles with centers C and D. If the st. line APB is parallel to CD, prove that AB = 2CD.
