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# Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

To do:

We have to prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:

Let two circles with centres $A$ and $A\'$ intersect each other at $B$ and $B\'$ respectively.

In $\triangle BAA’$ and $\triangle B\'AA’$

$AB = AB\'$ (Radii of circle with centre $A$)

$A’B = A’B\'$ (Radii of circle with centre $A\'$)

$AA’ = AA’$ (Common side)

Therefore, by $SSS$ congruency,

$\triangle BAA’ \cong \triangle B\'AA’$

This implies,

$\angle ABA\' = \angle AB\'A’$

From above,

The line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Hence proved.

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