Advertisements
Advertisements
प्रश्न
ABCD is a parallelogram. A circle through vertices A and B meets side BC at point P and side AD at point Q. Show that quadrilateral PCDQ is cyclic.
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
PQRS is a cyclic quadrilateral. Given ∠QPS = 73°, ∠PQS = 55° and ∠PSR = 82°, calculate:
1) ∠QRS
2) ∠RQS
3) ∠PRQ
In the following figure,
- if ∠BAD = 96°, find ∠BCD and ∠BFE.
- Prove that AD is parallel to FE.
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.
Calculate:
- ∠BEC
- ∠BED
In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L. Show that:
- ∠ONL + ∠OML = 180°
- ∠BAM + ∠BMA
- ALOB is a cyclic quadrilateral.
Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
ABCD is a cyclic quadrilateral, AB and DC are produced to meet in E. Prove that ΔEBC ≅ ΔEDA.
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.
If ∠ CAB = 34° , find : ∠ CQB
Prove that the angle bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (Provided they are not parallel) intersect at the right angle.
Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral ABCD is also cyclic.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
