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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 )


\ 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).


7.1 Introduction
7.2 Lines of circle
7.3 Arcs
7.4 Inscribed angels
7.5 Some properties od tangents, secants and chords
7.6 Chords and their arcs
7.7 Segments of chords secants and tangents
7.8 Lengths of arcs and area of sectors

Chapter 8

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