Advertisements
Advertisements
प्रश्न
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Advertisements
उत्तर
It is given that, ∠BCD = 100° and ∠ABD = 70°
As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.
\[\angle ADC + \angle ABC = 180° \]
\[ \Rightarrow 180° - 80° + 180° - x = 180°\]
\[ \Rightarrow x = 100°\]
In ΔABD we have,
\[\angle DAB + \angle ABD + \angle BDA = 180° \]
\[ \Rightarrow \angle BDA = 180° - 150° = 30° \]
Hence, `angle ABD = 30°`
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.
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.

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Prove that a cyclic parallelogram is a rectangle.
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.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
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.

In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.

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.
