Question

Prove that equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Answer 1

Hint: We use the theorem that the perpendicular drawn from the centre of a circle to any chord bisects the chord to have to get $BM=DN$. We then prove the congruence of the right angled triangles OBM and ODN which will give us $OM=ON$.

Complete step-by-step solution:
We see in the given figure that two chords of equal length $AB=CD$. We are also given in the centre O and perpendicular are dropped from O on AB as Om and on CD as ON. We are asked to prove AB and CD are equidistant from the centre O which means we have to prove $OM=ON$. \[\]
We know the theorem that the perpendicular (or radius) drawn from the centre of a circle to any chord bisects the chord. We use this theorem to deduce that OM will bisect AB to have
\[\begin{align}
  & AM=BM \\
 & \Rightarrow AM+BM=AB \\
 & \Rightarrow BM+BM=AB\left( \because AM=BM \right) \\
 & \Rightarrow 2BM=AB \\
 & \Rightarrow BM=\dfrac{1}{2}AB.....\left( 1 \right) \\
\end{align}\]
We use the theorem again to conclude that ON will bisect CD. So we have
 \[\begin{align}
  & CN=DN \\
 & \Rightarrow CN+DN=CD \\
 & \Rightarrow DN+DN=CD\left( \because CN=DN \right) \\
 & \Rightarrow 2DN=CD \\
 & \Rightarrow DN=\dfrac{1}{2}CD....\left( 2 \right) \\
\end{align}\]
Since we are given $AB=CD$ then we have
\[\begin{align}
  & \dfrac{1}{2}AB=\dfrac{1}{2}CD \\
 & \Rightarrow BM=DN\left( \text{from }\left( 1 \right)\text{ and }\left( 2 \right) \right) \\
\end{align}\]
We observe the right angled triangles OBM and ODN. We have already proved $BM=DN$ and the hypotenuse is $OB=OD$ because of $OB=OD$ are radii of the same circle. So hypotenuse-leg congruence we have
\[\Delta OBM\cong \Delta ODN\]
So the corresponding sides and angles are going to be equal. The only other pair of corresponding sides left are OM and ON. So we have $OM=ON$. Hence it is proved.\[\]

Note: We can proceed similarly to prove the result for two congruent circles (circles with equal radii) that equal chords are equidistant from centres. We note that when two right have equal hypotenuses and one other side (called leg) equal then they are congruent and it is called hypotenuse-leg congruence of right angled triangles.