Question

Chords which are equidistant from the center of congruent circles are congurent

Answer 1

Step-by-step explanation:

Let there is a circle having center O.

AB and Cd are the two chords and are equidistant from the centre.

So, form the figure,

OL = MO

Again, OL ⊥ AB and OM ⊥ CD

Now, join OA and OC.

Since OL ⊥ AB

So, AL = BL [since the perpendicular from the centre to a chord bisect the chord]

=> AL = AB/2 ..............1

Again, OM ⊥ CD

So, CM = DM [since the perpendicular from the centre to a chord bisect the chord]

=> CM = CD/2 .............2

Now, in ΔOAL and ΔOCM,

OA = OC {radii of the circle}

∠OLA = ∠OMC {each angle is equals to 90 degree}

OL = OM {given}

So, by RHS concruency, we get

ΔOAL ≅ ΔOCM

So, AL = CM

=> AB/2 = CD/2 {from equation 1 and 2}

=> AB = CD

Hence, chords equidistant from the centre are equal in length.