Theorem : The perpendicular from the centre of a circle to a chord bisects the chord.
Draw a circle. Let O be its centre. Draw a chord AB. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. So MA = MB.
Theorem : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Let AB be a chord of a circle with centre O and O is joined to the mid -point M of AB.
To prove: OM⊥ AB.
Proof: Join OA and OB
In triangles OAM and OBM,
OA =OB (radii of the same circle)
AM =BM (M is the midpoint of PQ)
OM =OM (common)
Therefore, ∆OAM ≅ ∆OBM (SSS congruence rule)
This gives , ∠OMA = ∠OMB = 90°
(Corresponding parts of congruent triangles are congruent).
Shaalaa.com | Theorem : The perpendicular from the centre of a circle to a chord bisects the chord.
Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball o Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer.