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

Let two circles having their centres as O and O’ intersect each other at point A and B respectively. Let us join OO’.

In ΔAOO’ and BOO’,
OA = OB ...(Radius of circle 1)
O’A = O’B ...(Radius of circle 2)
OO’ = OO’ ...(Common)
ΔAOO’ ≅ ΔBOO’ ...(By SSS congruence rule)
∠OAO’ = ∠OBO’ ...(By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
APPEARS IN
RELATED QUESTIONS
Prove that ‘Opposite angles of a cyclic quadrilateral are supplementary’.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Prove that a cyclic parallelogram is a rectangle.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.

In the figure, `square`ABCD is a cyclic quadrilateral. Seg AB is a diameter. If ∠ ADC = 120˚, complete the following activity to find measure of ∠ BAC.
`square` ABCD is a cyclic quadrilateral.
∴ ∠ ADC + ∠ ABC = 180°
∴ 120˚ + ∠ ABC = 180°
∴ ∠ ABC = ______
But ∠ ACB = ______ .......(angle in semicircle)
In Δ ABC,
∠ BAC + ∠ ACB + ∠ ABC = 180°
∴ ∠BAC + ______ = 180°
∴ ∠ BAC = ______
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.
