If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution:
Let AB and CD be the 2 equal chords. AB = CD. Let the chords intersect at point E. Join OE.
To prove AE = CE and BE = DE.
Draw perpendiculars from the center O to the chords. This Perpendicular bisects the chord AB at M and CD at N.
Thus, AM = MB = CN = DN ……(1)
In ∆OME and ∆ONE
∠M = ∠N = 90°
OE = OE
OM = ON (Equal chords are equidistant from the center.)
By RHS criteria, ∆OME and ∆ONE are congruent.
So by CPCT, ME = NE ….. (2)
We know that: CE = CN + NE and AE = AM + ME
From (1) and (2), it is evident CE = AE
DE = CD - CE and BE = AB - AE
AB and CD are equal, CE and AE are equal. So, DE and BE are also equal. It is proved corresponding segments of equal chords are equal.
☛ Check: Class 9 Maths NCERT Solutions Chapter 10
Video Solution:
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.4 Question 2
Summary:
If two equal chords of a circle intersect within the circle, then the segments of one chord are equal to corresponding segments of the other chord.
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