Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre.

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Solution

$∠ B O C = ∠ O A Z = 90$
corresponding $∠, O C ∥ Y Z$
$∠ C O A = ∠ X B O = 90^{o} (O C ∥ W X)$
$\Rightarrow ∠ B O C + ∠ C O A = 180^{o}$
$A O B$ us a straight line therefore $A B$ passes through $O$

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Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre.

∠BOC=∠OAZ=90corresponding ∠,OC∥YZ∠COA=∠XBO=90o(OC∥WX)⇒∠BOC+∠COA=180oAOB us a straight line therefore AB passes through O

$∠ B O C = ∠ O A Z = 90$
corresponding $∠, O C ∥ Y Z$
$∠ C O A = ∠ X B O = 90^{o} (O C ∥ W X)$
$\Rightarrow ∠ B O C + ∠ C O A = 180^{o}$
$A O B$ us a straight line therefore $A B$ passes through $O$