Question

abcd is a cyclic quadrilateral, o is the centre of the circle if angle bod = 160, find the angle bpd

Answer 1

BOD = 160*
so BAD = 160/2 = 80*
since ABCD is a cyclic quad. opp angles are supplementry
so BCD = 180-80 = 100*
therefore, BPD = 100* (since angle on the same segment are equal)

Hope this was the figure

Answer 2

Measure of angle BPD is 100° in cyclic Quadrilateral ABCD

Given:

ABCD is a cyclic quadrilateral

O is the center of the circle

∠BOD = 160°

To Find:

∠BPD

Solution:

An inscribed angle is half of a central angle that subtends the same arc

In a cyclic Quadrilateral measure of Sum of opposite angles is 180°

A Quadrilateral whose all vertices lies on a circle

Step 1:

Using inscribed angle theorem:

m∠BAD = (1/2)m∠BOD

Step 2:

Substitute m∠BOD = 160°

m∠BAD = (1/2)160°

=> m∠BAD = 80°

Step 3:

ABPD is a cyclic Quadrilateral hence sum of measures of opposite angles is 180°

(as A, B , P and D lies on the same circle hence ABPD is  a cyclic Quadrilateral )

m∠BAD + m∠BPD = 180°

Step 4:

Substitute m∠BAD = 80° in the equation and solve for m∠BPD

80° + m∠BPD = 180°

=> m∠BPD = 180° - 80°

=> m∠BPD = 100°

Measure of angle BPD is 100°

(Although your Question misses the figure , but relevant figure is attached)