Question

It two equal chords of a circle intersect within the circle. Prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer 1

Let us consider a circle with centre $O$ and $AB$ and $CD$ be the chords which intersect at $M$

According to the question, $AB=CD$.

We need to prove $\angle OME=\angle OMF$

Drop perpendiculars $OE$ and $OF$ on $AB$ and $CD$ respectively and join $OM$.

In triangles $OEM$ and $OFM$ ,

$OE=OF$ (equal chords of a circle are equidistant from the centre)

$OM=OM$ (common side)

$\angle OEM=\angle OFM$ (both equal to ${90}^o$)

By RHS criterion of congruence,
$△OEM\cong △OFM$

$\Rightarrow \angle OME=\angle OMF$ (by C.P.C.T.)