Two chords, AB and CD, are drawn in a circle with center O. The chords are at the same perpendicular distance from the center, with OP = OQ. Using properties of circles and Pythagoras' theorem, it is shown that OA = OC and AP = CQ. Since the perpendicular from the center bisects each chord, 1/2AB = 1/2CD, and therefore AB = CD. The conclusion is that two chords at the same perpendicular distance from the center are equal in length.

1. Good Morning...
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2. • O = CENTRE OF CIRCLE
• AB = DIAMETER OF CIRCLE
• CD = CHORD OF THE CIRCLE
O
CHORDS
A B
•
C D

3. EQUAL
CHORDS
•DIAMETER IS THE
LARGEST CHORD IN A
CIRCLE.
•AS THE CHORDS
MOVE AWAY FROM
THE CENTRE, THEIR
LENGTHS
DECREASE.

4. A

5. Two chords at the same perpendicular distance from the
centre. Prove that they are of same length. •O

6. •O
P
A
B
C
Let O be the centre and AB, CD be the
chords at the same perpendicular distance
OP and OQ from the centre.
We want to show that AB = CD
Join AO and CO.
Consider
OA = OC ( radii of the same circle )
OP = OQ ( given )
By Pythagoras theorem third sides are also
equal, ie , AP = CQ
½ AB = ½ CD ( perpendicular from the
centre bisects the chord)
Therefore, AB = CD
APO and CQO

7. “ TWO CHORDS AT THE
SAME PERPENDICULAR DISTANCE FROM THE
CENTRE
ARE EQUAL LENGTH.”