Prove that the line joining the mid-points of two equal chords of a circle subsent equal angles with the chord.

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Solution

Let, AB and CD be equal chords
Let, O be centre of the circle
Let, M & N be mid points of the chord.
Thus, $M B = A M$ & $C N = D N$
$\begin{matrix} ∴ O M = O N I n Δ O A B & Δ O C D O A = O D A B = C D O B = D C ∴ Δ O A B ≅ Δ O C D & (S S S c r i t e r i o n) ∴ ∠ A O B = ∠ C O D & (C P C T) \end{matrix}$

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Prove that the line joining the mid-points of two equal chords of a circle subsent equal angles with the chord.

Let, AB and CD be equal chordsLet, O be centre of the circleLet, M & N be mid points of the chord.Thus, MB=AM & CN=DN∴OM=ONInΔOAB&ΔOCDOA=ODAB=CDOB=DC∴ΔOAB≅ΔOCD(SSScriterion)∴∠AOB=∠COD(CPCT)