Question

✯ Prove that equal chords of a circle (or congruent circles) are equidistant from the centre (or centres).
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Answer 1

Given :-

A circle with center at O.

AB and CD are two equal chords of circle i.e. AB = CD & OX and OY are perpendiculars to AB & CD respectively.

To Prove :-

OX = OY

Proof :-

Since OX ⊥ AB

Perpendicular from the center to the chord, bisects the chord

$\sf AX = BX= \frac{AB}{2} \: \: \: \: \: \: \: ...(1)$

Proof :-

Since OY ⊥ CD

Perpendicular from the center to the chord, bisects the chord

$\sf CY = DY = \frac {CD} {2} \: \: \: \: \: \: ...(2)$

Now, given that

AB = CD

$\sf \frac{AB} {2} = \frac{CD} {2}$

AX = CY (From (1) and (2)) ...(3)

In △AOX and △COY

∠OXA = ∠OYC (Both 90°, given)

OA = OC (Radius)

AX = CY (From (1))

∴ ΔΑΟΧ ≅ ΔCOY (by R.H.S rule)

OX = OY (CPCT)

Hence, Proved.

Answer 2

Step-by-step explanation:

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