Advertisements
Advertisements
Question
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
Solution
(ii) It is given that BC || AD ,`angleDBC ` = 80° and `angle BAC` = 40°
We have to find `angleBCD`
`angleBAC = angleBDC `= 40° (Angle in the same segment are equal)
`angleDCB = `180° - (80° + 40° )
= 180° - 120°
= 60°
Hence, `angle BCD ` = 60°
APPEARS IN
RELATED QUESTIONS
Prove that "Opposite angles of a cyclic quadrilateral are supplementary".
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
If non-parallel sides of a trapezium are equal, prove that it is cyclic.