Circle theorems

Angle $CBD =$ angle $CAD = 35°$ (angles in the same segment are equal).

In triangle $BDE$: angle $BDE = 180 - 35 - 28 = 117°$.

Angle $ADC = 90°$ (angle in a semicircle), so angle $CDE = 180 - 90 = 90°$.

Angle $BDC = 117 - 90 = 27°$

$OA = OB$ (radii), so triangle $AOB$ is isosceles: $OBA = \frac{180 - x}{2} = 90 - \frac{x}{2}$

A tangent meets a radius at $90°$, so $OBC = 90°$

$ABC = 90 - \left(90 - \frac{x}{2}\right) = \frac{1}{2}x$

Angle $BAD = 132 \div 2 = 66^\circ$ (angle at the centre is twice the angle at the circumference).

Angle $BCD = 180 - 66 = 114^\circ$ (opposite angles of a cyclic quadrilateral add to 180°).

Angle $DCE = 180 - 114 = 66^\circ$ (angles on a straight line).

Angle $CDE = 180 - 66 - 16 = 98^\circ$ (angles in a triangle).

Angles $PQS$ and $PRS$ stand on the same arc $PS$, so angle $PQX$ = angle $SRX$ (angles in the same segment).

Angles $QPR$ and $QSR$ stand on arc $QR$, so angle $QPX$ = angle $RSX$ (angles in the same segment).

Angle $PXQ$ = angle $SXR$ (vertically opposite).

All three pairs of angles are equal, so the triangles are similar.

Let angle $BCD = x$, so angle $EAB = 2x$.

Opposite angles of a cyclic quadrilateral add to $180^\circ$, so angle $BDE = 180 - 2x$, and angle $BDC = 180 - (180 - 2x) = 2x$ (angles on a straight line).

Triangle $BCD$: $60 + x + 2x = 180$, so $3x = 120$ and $x = 40^\circ$

Opposite angles of a cyclic quadrilateral add to 180°, so $BCD = 180 \div 4 = 45°$ and $BAD = 135°$.

Alternate segment theorem: angle $SBD = BCD = 45°$.

$SBA = SBD - ABD = 45 - 20 = 25°$.

$BDEF$ is a cyclic quadrilateral, so angle $DBF = 180^\circ - 100^\circ = 80^\circ$ (opposite angles of a cyclic quadrilateral add to $180^\circ$).

In triangle $BDF$: angle $BFD = 180^\circ - 80^\circ - 40^\circ = 60^\circ$ (angles in a triangle add to $180^\circ$).

Angle $ABD$ = angle $BFD = 60^\circ$ (alternate segment theorem).

Triangle ABD is isosceles, so $\angle BAD = \angle BDA = \frac{180 - 40}{2} = 70^\circ$ (base angles of an isosceles triangle are equal; angles in a triangle sum to 180°).

$\angle BCD = 180 - 70 = 110^\circ$ (opposite angles of a cyclic quadrilateral sum to 180°).

Triangle BCD is isosceles: $\angle CDB = \frac{180 - 110}{2} = 35^\circ$.

$\angle ADE = 180 - 70 - 35 = 75^\circ$ (angles on a straight line sum to 180°).

The radius meets the tangent at 90°, so angle $OAD = 90°$.

Angle $AOD = 180 - 90 - 32 = 58°$, so angle $AOB = 180 - 58 = 122°$.

$OA = OB$ (radii), so angle $OAB = (180 - 122) \div 2 = 29°$.

Angle $CAB = 90 - 29 = 61°$

$OA = OB$ (radii), so triangle $OAB$ is isosceles: angle $BAO = x$ (base angles of an isosceles triangle are equal).

Angle $OBC = 90^\circ$ (a tangent meets a radius at $90^\circ$), so angle $ABC = 90 + x$.

In triangle $ABC$ the angles add to $180^\circ$: angle $ACB = 180 - x - (90 + x) = 90 - 2x$

Tangent meets radius at 90°, so angle OAB $= 90° - 56° = 34°$.

OA = OB, so angle OBA $= 34°$ and angle ABC $= 34° + 35° = 69°$.

Alternate segment theorem: angle ACB $= 56°$.

Angle CAB $= 180° - 69° - 56° = 55°$

Angle CAO $= 55° - 34° = 21°$

(a) Angle $ODE = 90°$ (tangent meets radius at right angles), so angle $BDE = 90 + x$.

By the alternate segment theorem, angle $BAD =$ angle $BDE$, so $y = 90 + x$.

Therefore $y - x = 90$.

(b) No. $y$ is an angle of triangle $ABD$, so it is less than 180°; it cannot be 200.

Alternate segment theorem: angle $BAC =$ angle $BCD = 40°$

Angle $OAB = 3 \times$ angle $OAC$ and they add to $40°$, so angle $OAC = 10°$ and angle $OAB = 30°$

Triangle $OAB$ is isosceles, so angle $AOB = 180 - 30 - 30 = 120°$

Angle at the centre is twice the angle at the circumference: angle $ACB = 60°$

Angle $ACD = 60 + 40 = 100°$

Join OA, OB and OP.

$\angle OAP = \angle OBP = 90^\circ$ (a tangent is perpendicular to the radius).

$OA = OB$ (both radii) and $OP$ is common.

So triangles $OAP$ and $OBP$ are congruent (RHS), hence $AP = BP$. (Or: $AP^2 = OP^2 - OA^2 = OP^2 - OB^2 = BP^2$.)

Angle $BAD = 64 \div 2 = 32^\circ$ (angle at the centre is twice the angle at the circumference).

Let $AD$ meet $OB$ at $P$. In triangle $ABP$: angle $APB = 180 - 51 - 32 = 97^\circ$, so angle $OPD = 97^\circ$ (vertically opposite).

In triangle $OPD$: angle $ODA = 180 - 64 - 97 = 19^\circ$

Angle $ODC = 90^\circ$ (tangent is perpendicular to the radius), so angle $ADC = 90 - 19 = 71^\circ$

Triangle $AED$ is equilateral, so each angle is $60°$.

$\angle ACB = \angle ADB = 60°$ and $\angle DBC = \angle DAC = 60°$ (angles in the same segment).

So triangle $BEC$ is equilateral and $\angle ACB = \angle DBC$.

$AC = AE + EC = DE + EB = DB$ (sides of equilateral triangles).

$BC$ is common to both triangles.

So triangles $ABC$ and $DCB$ are congruent (SAS).

Let angle $CAB = x$.

$AORC$ is cyclic, so angle $CRO = 180 - x$ (opposite angles of a cyclic quadrilateral add up to $180°$).

Angle $ORB = 180 - (180 - x) = x$ (angles on a straight line add up to $180°$).

$OR = OB$ (both radii), so triangle $ORB$ is isosceles and angle $OBR = x$ (base angles are equal).

So angle $ABC = x =$ angle $CAB$.

Join $AO$ and extend it to $D$. Let $\angle OAB = x$ and $\angle OAC = y$.

$OA = OB$ (radii), so $\angle OBA = x$ (base angles of an isosceles triangle).

Exterior angle of triangle $AOB$: $\angle BOD = x + x = 2x$

Similarly $\angle COD = 2y$

$\angle BOC = 2x + 2y = 2(x + y) = 2 \times \angle BAC$