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
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
In the figure m(arc LN) = 110°,
m(arc PQ) = 50° then complete the following activity to find ∠LMN.
∠ LMN = `1/2` [m(arc LN) - _______]
∴ ∠ LMN = `1/2` [_________ - 50°]
∴ ∠ LMN = `1/2` × _________
∴ ∠ LMN = __________
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
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.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
