Question

Prove that the line segment joining the midpoints of two equal chords of a circle subtends equal angles with the chord.

Answer 1

Here, M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O.
We have to prove that
∠BMN = ∠CNM
∠AMN = ∠DNM

Join ON, OM and NM
∴ ∠OMA = ∠OMB = 90°     
∠ OND = ∠ONC = 90°                ....(i)(Line joining the centre and midpoint of a chord is perpendicular to the chord)

Since, AB = CD ⇒ OM = ON
∴ In ΔOMN,  ∠OMN = ∠ONM     ...(ii)

(i) ∠OMB = ∠ONC              ...[ Using (i) and (ii) ]
∠OMN = ∠ONM
⇒ ∠OMB - ∠OMN = ∠ONC - ∠ONM
⇒ ∠BMN = ∠CNM

(ii)  ∠OMA = ∠OND
∠OMN = ∠ONM            ...[ Using (i) and (ii) ]
⇒ ∠OMA + ∠OMN = ∠OND + ∠ONM    

⇒ ∠AMN = ∠DNM
Hence proved.