#### Question

In the figure given below, an AD is a diameter. O is the centre of the circle

AD is parallel to BC and `angle CBD` = 32°. Find

1) `angle OBD`

2) `angle AOB`

3) `angle BED`

#### Solution

1) AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal

`=> angle ODB = angle CBD = 32^@` .... (Alternate angles)

In ΔOBD

OD = OB ....(Radii of the same circle)

`=> angle ODB = angle OBD = 32^@`

2) AD is parallel to BC, i.e., AO is parallel to BC and OB is transversal

`=> angle AOB = angle OBC` .... (Alternate angles)

`=> angle OBC = angle OBD + angle DBC`

`=> angle OBC = 32^@ + 32^@`

`=> angle OBC = 64^@`

`=> angle AOB = 64^@`

3) In ΔOAB,

OA = OB . . . .(Radii of the same circle)

`=> angle OAB = angleOBA` = x(say)

`angle OAB + angle OBA + angle AOB = 180^@`

`=> x + x+ 64^@ = 180^@`

`=> 2x = 180^@ -64^@`

`=> 2x = 116^@`

`=> x = 58^@`

`=> angle OAB = 58^@`

`i.e angle DAB = 58^@`

`=> angle DAB = angle BED = 58^@` ...(Angles inscribed in the same arc are equal)