Advertisements
Advertisements
प्रश्न
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Advertisements
उत्तर

To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.
Proof:
We know that the perpendicular bisector of every chord of a circle always passes through the centre.
Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.
Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.
APPEARS IN
संबंधित प्रश्न
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
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.

In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.

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.
