The lines ABC and CDE are tangents to the circle at B and D respectively. F, G and H are points on the circle and the centre of the circle is O. The straight lines BD and HF intersect at J. It is given that ∠ACE = 62° and ∠DHF = 40°. Find, giving reasons for each answer,

1. Identify equal angles subtended by the same chord

$\angle DHF$ and $\angle FGD$ are angles at the circumference standing on the same chord $DF$.

Hence,
$\angle FGD = \angle DHF = 40^\circ$
$\text{(angles in the same segment)}$

1. Apply the tangent-radius property at the point of contact $D$

Since $CDE$ is a tangent to the circle at $D$ and $OD$ is a radius,

$\angle CDO = 90^\circ$.

Reason: tangent is perpendicular to the radius at the point of contact.

1. Find the central angle subtended by the two points of contact $B$ and $D$

Since $OB \perp CB$ and $OD \perp CD$,
$\angle OBC=90^\circ,\quad \angle ODC=90^\circ.$
In quadrilateral $BCDO$,
$\angle BOD=360^\circ-90^\circ-90^\circ-62^\circ=118^\circ.$

2. Find the central angle subtended by chord $DF$

$\angle DHF=40^\circ$ is an angle at the circumference subtended by chord $DF$.

Therefore,
$\angle DOF=2\times 40^\circ=80^\circ$
(angle at the centre is twice the angle at the circumference).

3. Subtract central angles to find $\angle BOF$

Since $F$ lies on the arc between $B$ and $D$,
$\angle BOF=\angle BOD-\angle DOF=118^\circ-80^\circ=38^\circ.$