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Figure 7.19
Thus if two chords are equal in measure they are equidistant from the center of the circle.
The converse of this theorem is that if two chords are equidistant from the center of the circle, they are equal in measure.
As shown in figure 7.20 if seg. HI and seg. JK are two chords equidistant from the center of the circle, they are equal in length.
Figure 7.20
To prove that seg.HI @
seg.JK join OI and OK.
Consider D
OIP and D OKQ, ( both are
right triangles) .
seg.OI @
seg.OK, ( both are radii of the same circle).
seg.OP @
seg.OQ (given that chords are equidistant from the center O).
\
D
OIP @
D OKQ (H.S.)
\
seg.PI @ seg.QK (corresponding
sides of congruent triangles are congruent).
Also it is known that the perpendicular from
the center bisects the chord. Therefore, seg. HI @
seg JK.
Example 1
AB and CD are chords in a circle with center O.
l
(seg.AB ) = l
(seg.CD) = 3.5 cm and m Ð
COD = 950.
Find m arc AB.
Solution:
950
m arc AB = m Ð
AOB
Since D
AOB @
D
COD by SSS m Ð
AOB = m Ð COD.
Example 2
PQ is a chord of a circle with center O. Seg.OR
is a radius intersecting PQ at right angles at point T. If
l (PT) = 1.5 cm and m arc PQ
= 800,
find l (PQ) and
m arc PR.
Solution:
l (PQ) = 3
m (arc PR) = 400
Seg.OT is perpendicular to PQ and therefore bisects PQ at T.
\ l
(.PQ) = 2 l (PT)
Seg.OR bisects arc PQ. \
m (arc PR) = 
m (arc PQ)
Example 3
Seg HI and seg. JK are chords of equal measure in a circle with center O. If the distance between O and seg. HI is 10 cm find the length of the perpendicular from O onto seg.JK.
Solution:
10 cm.
Chords of equal measure are equidistant from the center.
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