Advertisements
Advertisements
प्रश्न
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
Advertisements
उत्तर
Given: ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q.
To prove: Points P, Q, C and D are con-cyclic.

Construction: Join PQ
Proof: ∠1 = ∠A ...[Exterior angle property of cyclic quadrilateral]
But ∠A = ∠C ...[Opposite angles of a parallelogram]
∴ ∠1 = ∠C ...(i)
But ∠C + ∠D = 180° ...[Sum of cointerior angles on same side is 180°]
⇒ ∠1 + ∠D = 180° ...[From equation (i)]
Thus, the quadrilateral QCDP is cyclic.
So, the points P, Q, C and D are con-cyclic.
Hence proved.
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 the non-parallel sides of a trapezium are equal, prove that it is cyclic.
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.

Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
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.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
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.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
