A circle with center B has chords DC and FA which are equidistant from B.
A chord is a straight line drawn from a point on the circumference of a circle to another point on the circumference, without passing through the center of the circle.
From the given question, a circle center B has two chords FA and DC that are equidistant from the center of the circle.
The center of a circle is the point where we place the tip of our compass while drawing a circle.
The distant of FA from B is BG and the distance of DC from B is BE. But BG = BE, which implies that the two chords have equal distant from the center of the circle.
Therefore line BG from center B intersects FA at G, and line BE from center B intersects FA at E.
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It can be concluded that chords DC and FA will be equal in length.
A circle is a locus of a point that is always equidistant from a fixed point called the center of the circle.
Join B to D and B to A.
In triangle BGD and triangle BEA
∠BGD = ∠BEA = 90°
BG=BE (Given)
BD=BA (both are radius)
So, ΔBGD≅ΔBEA
So, GD =EA......(1)
We know the perpendicular drawn from the center of the circle to the chord bisects the chord.
So, GD = GC =1/2DC
similarly, EA = EF = 1/2FA
Put these values in (1)
1/2DC = 1/2FA
So, DC=FA
Hence, it can be concluded that chords DC and FA will be equal in length.
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