Question

Prove that the equal chords of two congruent circles subtend equal angles at their respective centres.

Answer 1

Hint: Use the fact that the radii of two congruent circles are equal and hence prove that OA = O’A’ and OB = O’B’. Use the fact that since the chords are equal AB = A’B’ and hence prove that the triangle ABC and A’B’C’ are congruent and hence prove that $\mathrm{\angle }AOC=\mathrm{\angle }{A}^{\prime }{O}^{\prime }{C}^{\prime }$. Hence prove that the equal chords of congruent circles subtend equal angles at the centres of their corresponding circles.

Complete step by step answer:

Given: Two circles with centre O and O’ have equal radii. AB is the chord of the circle with centre O and A’B’ is a chord of the circle with centre O’.
To prove $\mathrm{\angle }AOB=\mathrm{\angle }{A}^{\prime }{O}^{\prime }{B}^{\prime }$
Proof:
Since the circle have equal radii, we have
OA = O’A’ and OB = O’B’
Now, in triangle AOB and A’O’B’, we have
AO = A’O’ (proved above)
OB = O’B’ (Proved above)
AB = A’B’ (Given).
Hence by S.S.S congruence criterion, we have
$\mathrm{\Delta }AOB\cong \mathrm{\Delta }{A}^{\prime }{O}^{\prime }{B}^{\prime }$
Hence, we have
$\mathrm{\angle }AOB=\mathrm{\angle }{A}^{\prime }{O}^{\prime }{B}^{\prime }$ (Corresponding parts of congruent triangles)
Hence, equal chords of two congruent circles subtend equal angles at their respective centres.
Hence, proved.

Note: Alternative Solution: Using Sine rule.
We know that if R is the radius of circumcentre of a triangle ABC, then ${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=2R$

Consider two points C and C’ on the alternate segments as shown in the diagram above.
By sine rule, we have
$\begin{array}{rl}& {\displaystyle \frac{AB}{\sin C}}=2R\\ & \Rightarrow AB=2R\sin C\end{array}$
Similarly, we have
$A^{\prime }{B}^{\prime }=2R\sin C^{\prime }$
Since AB = A’B’, we have
$\begin{array}{rl}& 2R\sin C=2R\sin C^{\prime }\\ & \Rightarrow \sin C=\sin C^{\prime }\\ & \Rightarrow C=C^{\prime }\end{array}$
We know that the angle subtended in the alternate segment is half the angle subtended at the centre.
Hence, we have
$\mathrm{\angle }AOB=2\mathrm{\angle }C$ and $\mathrm{\angle }{A}^{\prime }{O}^{\prime }{B}^{\prime }=2\mathrm{\angle }{C}^{\prime }$
Since $\mathrm{\angle }C=\mathrm{\angle }{C}^{\prime }$, we have
$\mathrm{\angle }AOB=\mathrm{\angle }{A}^{\prime }{O}^{\prime }{B}^{\prime }$
Q.E.D