Question

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the center makes equal angles with the chords.

Answer 1

Given: Two equal chords which are intersecting at point in the circle

Construction :

1. Draw two equal chords $AB$ and $CD$ in a circle which intersect at point $E$
2. Draw a diameter $PQ$ such that it passes through the point $E$
3. Draw two perpendiculars are drawn as $OM\perp AB$ and $ON\perp D$. Now, join $OE$.

To Prove

$\angle BEQ=\angle CEQ$

Proof:

$OM=ON$[Since the equal chords are always equidistant from the center]

$OE=OE$[common side]

$\angle OME=\angle ONE=90°$ [These are the perpendiculars]

So, by the RHS congruency criterion, $\Delta OEM\cong \Delta OEN.$

So, by the CPCT rule, $\angle MEO=\angle NEO$

$\therefore \angle BEQ=\angle CEQ$

Hence it is proven that the line joining the point of intersection of two equal chords to the center makes equal angles with the chords.