Question

o, b, d and c are the vertices of a rhombus and a, b, d and c lie on the circle with centre o as shown in figure find angle boc angle obc angle bac and angle bdc

Answer 1

In triangle OBE and OCE,
OB=OC (radii)
BE=EC (Diagnols)
OE=OE (common)
Hence both triangles are congruent
By CPCT, angle BOE=angle COE
As both are equilateral triangle Hence
angle BOE=angle COE=60°
Through this angle BOC =120°
Now angle 2BAC=angle BOC (central angle is double of inscribed angle)
Hence angle BAC=60°
AsABCD is a cyclic quadrilateral Hence sum of opposite sides should be 180°
Through this Angle BDC=120°
Hope this helps you

Answer 2

Answer:

∠BOC = 120°

∠OBC = 30°

∠BAC = 60°

∠BDC = 120°

Step-by-step explanation:

Here, from the given figure,

In Triangle OBD and Triangle OCD,

OB = OC ....... ∵ radii of the circle

BD = DC .... ∵ Diagonals of rhombus

and both triangles have OD in common.

As all sides of both triangles are congruent (equal/similar), hence Both Triangles are congruent.

Hence by CPCT,

∠BOD = ∠COD

CPCT stands for Corresponding parts of congruent triangles are congruent.

Hence both triangles are equilateral.

Hence,

∠BOD = ∠COD = 60°

Now as ∠BOC = ∠BOD + ∠COD = 60° + 60° = 120°

∴ ∠BOC = 120°

Now, as the Central angle is twice the inscribed angle and ∠BOC is the central angle and ∠BAC is the inscribed angle,

∴ 2(∠BAC) = ∠BOC

∴∠BAC = 60°

Now, as OD and BC are the diagonals of the Rhombus, hence they intersect each other perpendicular forming a right angle at intersecting point F.

Hence, in Triangle BFD,

∠BFD = 90° and

from above ∠OBD = 60°........  ∵ (angle of equilateral triangle OBE)

∴ ∠OBC = 30° ...... (sum of all angles of a triangle is 180°)

Now, as ABCD is cyclic quadrilateral, hence sum of opposite angles is 180°

By this, ∠BAC + ∠BDC = 180°

∴ ∠BDC = 180° - 60° = 120°

∴∠BDC = 120°