The points A, B, E and D lie on the circle with centre O. BEC is a straight line and BF is a tangent to the circle at point B. It is given that AC is parallel to BF, angle ABC = 90° and angle ACB = 34°. Stating reasons clearly, calculate

1. Find the angle between chord $BE$ and the tangent $BF$.

Since $B,E,C$ are collinear, $BE$ lies on the same straight line as $BC$.

Since $AC \parallel BF$, the angle between $BE$ and $BF$ is equal to $\angle ACB$.

Hence,
$\angle EBF = 34^\circ$
(alternate/corresponding angles, since $AC \parallel BF$).

2. Use radius-tangent perpendicularity to find $\angle OBE$.

$OB$ is a radius to the point of tangency $B$, so
$\angle OBF = 90^\circ$
(radius is perpendicular to tangent).

Therefore,
$\angle OBE = 90^\circ - 34^\circ = 56^\circ.$

1. Find the angle at $A$ in triangle $ABC$.

In $\triangle ABC$,
$\angle BAC=180^\circ-90^\circ-34^\circ=56^\circ.$

2. Use the cyclic quadrilateral property to find $\angle DEB$.

Since $A, B, E$ and $D$ lie on the same circle, $ABED$ is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary, so
$\angle DEB=180^\circ-\angle DAB.$

As $A, D, C$ are collinear, $\angle DAB=\angle CAB=56^\circ$.

Hence
$\angle DEB=180^\circ-56^\circ=124^\circ.$

1. Find the central angle subtended by chord $BD$.

From part (b), $\angle DEB=124^\circ$, so the opposite angle in the cyclic quadrilateral is
$\angle BAD=180^\circ-124^\circ=56^\circ.$
Since $\angle BOD$ and $\angle BAD$ stand on the same chord $BD$,
$\angle BOD=2\times 56^\circ=112^\circ.$
Reason: angle at the centre is twice the angle at the circumference.