7.6 Chords and their arcs
Theorem: If in any circle two chords are equal in length then the measures of their corresponding minor arcs are same.
As shown in figure 8.16 AB and CD are congruent
chords. Therefore according to the theorem stated above m (arc AB)
= m (arc CD) or m Ð
AOB = m Ð COD.
Figure 7.16
To prove this, join A, B, C & D withO.
Consider D
AOB and D COD
seg.AO @
seg.CO and seg.OB @ seg.OD
(radii of a circle are always congruent).
seg.AB @
seg.CD (given)
\
D
AOB @
D COD ( S S S )
\
Ð
AOB @
Ð COD (corresponding
angles of congruent triangles are congruent).
Hence arce AB @
arc CD.
Conversely it can also be proved that in the same circle or congruent circles, congruent arcs have their chords congruent.
Figure 7.17
In figure 7.17 if arc LM @
arc PQ then seg. LM @ seg.
PQ.
To prove this, join L, M, P & Q to O.
Consider D
LOM and D POQ
seg. OL @
seg. OP and seg. MO @ seg
QO ( as all are radii ).
Ð
LOM @
Ð POQ (given )
\
D
LOM @
D POQ (SAS)
\
seg. LM @ seg PQ (corresponding
sides of congruent triangles are congruent).
Theorem: The perpendicular from the center of a circle to a chord of the circle bisects the chord.
Figure 7.18
In figure 7.18, XY is the chord of a circle with center O. Seg.OP is the perpendicular from the center to the chord. According to the theorem given above seg XP = seg. YP.
To prove this, join OX and OY
Consider D
OXP and D OXY
Both are right triangles.
hypotenuse seg. OX @ hypotenuse
seg. OY
( both are radii of the circle )
seg. OP @ seg OP (same
side)
\
D
OXP @
D OYP (H.S.)
\
seg XP @ seg. YP (corresponding
sides of congruent triangles are congruent).
\ P is the midpoint of seg.XY.
Hence seg.OP which is the perpendicular from the center to the chord seg.XY bisects the chord seg.XY.
Now consider two chords of equal length in the same circle. Their distance from the center of the circle is same.
In figure 7.19 seg. PQ and seg. RS are two
chords in the circle with center O such that l
(seg.PQ) = l (seg.RS).
seg.OX and seg. OY are the perpendicular distances from O to seg.PQ and seg.RS respectively.
To prove that l
(seg.OX) = l (seg.OY)
join O to P and R.
Since perpendicular from the center to the
chord bisects the chord l
( seg.PX) = l (seg.RY).
Now consider D
POX and D ROY
seg. PX @ seg.RY
seg.OP @ seg.OR. Both being
radii onto the circle.
\
D
POX @
D ROY (by H.S.)
\
seg.OX @ seg.OY (corresponding
sides of congruent triangles are congruent).
