A, B, C, D and E are points on the circle, centre O. QP and QR are tangents to the circle at B and E respectively. AC intersects BD at O and QAOC is a straight line. Angle BOA = 50°. Find, giving reasons for each answer,

1. Identify the central angle subtended by chord $CD$.

Since $AC$ and $BD$ pass through the centre $O$, $AC$ and $BD$ are diameters.

Hence $A, O, C$ are collinear and $B, O, D$ are collinear, so
$\angle COD = \angle AOB = 50^\circ$
vertically opposite angles.

2. Use the angle at the circumference theorem to find $\angle CED$.

$\angle CED$ is the angle at the circumference standing on the same arc $CD$ as the central angle $\angle COD$.

Therefore,
$\angle CED = \frac{1}{2}\angle COD = \frac{1}{2}(50^\circ)=25^\circ.$

1. Find the angle between $QB$ and $QO$ using the right angle between a radius and tangent.

Since $QB$ is a tangent at $B$, $OB \perp QB$.

Also, $Q, A, O, C$ are collinear, so
$\angle BOQ = \angle BOA = 50^\circ.$

In $\triangle QOB$,
$\angle BQO = 180^\circ - 90^\circ - 50^\circ = 40^\circ.$

2. Use the tangent angle-bisector property to find $\angle BQE$.

Tangents $QB$ and $QE$ are drawn from the same external point $Q$, so $OQ$ bisects $\angle BQE$.

Hence
$\angle BQE = 2\angle BQO = 2(40^\circ)=80^\circ.$

1. Find the central angle $\angle AOE$.

Since $QB$ and $QE$ are tangents from the same external point, $QO$ bisects $\angle BOE$.

Also $Q,A,O,C$ are collinear, so $OQ$ is the same line as $OA$.

Hence
$\angle AOE = \angle QOE = \angle BOQ = 50^\circ$.

2. Compare the angles made by $AE$ and $BD$ with the straight line $AC$.

In $\triangle AOE$, $OA=OE$ since both are radii.

So
$\angle OAE=\angle AEO=\frac{180^\circ-50^\circ}{2}=65^\circ.$

Since $A,O,C$ are collinear, $\angle CAE=\angle OAE=65^\circ$.

But $BD$ is the same straight line as $BO$, so the angle between $BD$ and $AC$ is
$\angle BOA=50^\circ.$

Since $65^\circ \ne 50^\circ$, $BD$ is not parallel to $AE$.