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
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
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.
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 = ______
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
