Advertisements
Advertisements
प्रश्न
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.

Advertisements
उत्तर

Join chords AP and DQ.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) ... (1)
For chord DQ,
∠DBQ = ∠QCD (Angles in the same segment) ... (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠PBA = ∠DBQ (Vertically opposite angles) ... (3)
From equations (1), (2), and (3), we obtain
∠ACP = ∠QCD
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.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
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.
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.

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 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 = ______
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.

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°.
