Question

Prove that if chords of congruent circles subtend equal angles their centres, then the chords are equal.

Answer 1

Take two congruent circles, and let their centres be $O_1$ and $O_2$

Let, the chords making equal angles at centres be $AB$ and $CD$ respectively for two circles.

Let the radius of both circles be $r$.

We know that $O_{1}A=O_{1}B=O_{2}C=O_{2}D$.

Since, $O_{1}A=O_{1}B=O_{2}C=O_{2}D$, we get,

$\angle O_{1}AB=\angle O_{1}BA\angle O_{2}CD=\angle O_{2}DC$.

So, we get $\angle A{O}_{1}B=\angle C{O}_{2}D$.

So, by $SAS$ congruency, we get triangle $A{O}_{1}B$ is congruent to triangle $C{O}_{2}D$.

$\therefore AB=CD$, which implies that the length of chords are equal.