Mathematics
The line joining the mid-points of two chords of a circle passes through its center. Prove that the chords are parallel.
Answer
Since, the straight line drawn from the center of a circle to bisect a chord is perpendicular to the chord,
∴ OM ⊥ AB and ON ⊥ CD.
So,
⇒ ∠OMA = ∠OMB = 90° and ∠ONC = ∠OND = 90°
Since,
⇒ ∠OMA = ∠OND = 90° (Alternate angles are equal)
⇒ ∠OMB = ∠ONC = 90° (Alternate angles are equal)
Hence, proved that AB || CD.
Related Questions
In the following figure, OABC is a square. A circle is drawn with O as center which meets OC at P and OA at Q. Prove that :
(i) △ OPA ≅ △ OQC
(ii) △ BPC ≅ △ BQA
The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles be 50 cm and 34 cm, calculate the distance between their centers.
In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centers of the circles. Show that :
(i) AB = CD
(ii) AC = BD
In the given figure, arc APB = arc CQD, then :
AB = CD
AB > CD
AB < CD
none of the above