Question

If two chords of a circle are equidistant from the center of the circle then they are
(a)Equal to each other
(b)Not equal to each other
(c)Intersect each other
(d)None of these

Answer 1

Hint: Draw a circle, let AB and CD be the chords and E the center. Prove that the triangle, AEB and CED are congruent. Draw $$EF\mathrm{\perp }AB$$ & $$EG\mathrm{\perp }CD$$. Prove that the triangle AEF, BEF, CEG, DEG are similar to each other. Then by CPCT prove that EF = EG i.e. the chords are equidistant from the center.

Complete step-by-step answer:
Let's consider AB and CD are the two chords of the circle. Let E be the center of the circle. It is said that the two chords are equidistant from the center of the circle.

Hence we can say that AB and CD are equidistant from point E.
Now let us join EA, EB, EC and ED. Thus we get two triangles $$\mathrm{\Delta }AEB$$ and $$\mathrm{\Delta }CED$$. Let us check if they are congruent.
EA = EB = EC = ED, all there are the radius of the circle. Thus all the radii in a triangle are the same. Hence by SSS congruence we can say that three sides of $$\mathrm{\Delta }AEB$$ is equal to three sides of $$\mathrm{\Delta }CED$$. Hence we can say that $$\mathrm{\Delta }AEB$$ and $$\mathrm{\Delta }CED$$ are similar.
$$\therefore \mathrm{\Delta }AEB\cong \mathrm{\Delta }CED$$
Now let us draw the altitudes of the isosceles triangles. Let EF be the altitude in $$\mathrm{\Delta }EAB$$ and EG be the altitude in $$\mathrm{\Delta }CED$$. Thus we can say that, $$EF\mathrm{\perp }AB$$ & $$EG\mathrm{\perp }CD$$.
We know that as $$EF\mathrm{\perp }AB$$, the angle made will be $${90}^{\circ }$$.
i.e. $$\mathrm{\angle }EFA=\mathrm{\angle }EFB={90}^{\circ }$$, they are right angles.
Similarly, as $$EG\mathrm{\perp }CD$$, the angle made will be right angles,
i.e. $$\mathrm{\angle }EGC=\mathrm{\angle }EGD={90}^{\circ }$$
Thus we can say that,
$$\mathrm{\angle }EFA=\mathrm{\angle }EFB=\mathrm{\angle }EGC=\mathrm{\angle }EGD$$, all right angles are equal.
Thus we can say from the figure, $$\mathrm{\Delta }AEF,\mathrm{\Delta }BEF,\mathrm{\Delta }CGE$$ and $$\mathrm{\Delta }GDE$$ are equal as per AAS criteria, which is two angles and one side are equal i.e.
In $$\mathrm{\Delta }AEF$$ and $$\mathrm{\Delta }BEF$$
EA = EB [radius of the circle]
$$\mathrm{\angle }EFA=\mathrm{\angle }EFB={90}^{\circ }$$ [right angles]
$$\mathrm{\angle }AEF=\mathrm{\angle }BEF$$ [angle subtended by the radius of the circle is equal]
$$\therefore \mathrm{\Delta }AEF\cong \mathrm{\Delta }BEF$$ [similar by AAS criteria]
Similarly, we can say that,
$$\mathrm{\Delta }AEF\cong \mathrm{\Delta }BEF\cong \mathrm{\Delta }CEG\cong \mathrm{\Delta }DEG$$
Hence by CPCT [corresponding parts of the congruent triangle] we can say that, EF = EG.
Hence EF and EG are equal. AB and CD are equidistant from the center of the center E.
Thus if two chords are equidistant from the center of a circle, then the chords are equal.
Hence, if two chords are equidistant from the center the center of the circle then they are equal to each other.
$$\therefore$$ Option (a) is the correct answer.

Note: We have been told that EF = EG.
$$\therefore$$ In $$\mathrm{\Delta }AEF$$, where AE = r, applying Pythagoras theorem, we get
$$\Rightarrow$$$$A{E}^2=E{F}^2+A{F}^2\Rightarrow A{F}^2=r^2-E{F}^2$$
In $$\mathrm{\Delta }CGE$$, where CE = r, EF = EG, applying Pythagoras theorem, we get
$$\Rightarrow C{E}^2=E{G}^2+C{G}^2\Rightarrow C{G}^2=r^2-E{G}^2=r^2-E{F}^2$$
$$\Rightarrow A{F}^2=C{G}^2$$ i.e. $$AF=CG$$ i.e. AB = CD