Advertisements
Advertisements
प्रश्न
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove that ‘Opposite angles of a cyclic quadrilateral are supplementary’.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.
