80^\circ - 2y) = 2x + 2y $$ 7. The angle at the circumference is $\angle PQR = x + y$. 8. Therefore the angle at the centre is $2(x+y)$, which is twice the angle at the circumference. If the question says “do not use circle theorems”, you must not write “angle at centre is twice angle at circumference”. Prove it from radii and triangle angle facts instead. ## 3. Angle in a semicircle A semicircle is half a circle. If a triangle has one side as the diameter of a circle, the angle opposite the diameter is a right angle. Points $A$, $B$ and $C$ lie on a circle with centre $O$, and $AC$ is a diameter. Prove that $\angle ABC = 90^\circ$ without using circle theorems. 1. Join $OB$. Let $\angle ABO = x$ and $\angle OBC = y$. 2. Since $OA = OB$, triangle $AOB$ is isosceles, so $\angle BAO = x$. 3. Since $OB = OC$, triangle $BOC$ is isosceles, so $\angle BCO = y$. 4. In triangle $AOB$, $\angle AOB = 180^\circ - 2x$. 5. In triangle $BOC$, $\angle BOC = 180^\circ - 2y$. 6. Since $AOC$ is a straight line, $\angle AOB + \angle BOC = 180^\circ$. 7. So $(180^\circ - 2x) + (180^\circ - 2y) = 180^\circ$, giving $x+y=90^\circ$. 8. Since $\angle ABC = x+y$, $\angle ABC = 90^\circ$. If a line passes through the centre and both ends are on the circle, it is a diameter. The angle opposite it on the circumference will be 90°. ## 4. Angles in the same segment A **segment** is the region between a chord and an arc. Angles in the same segment stand on the same chord and sit on the same side of it. Points $A$, $B$, $C$ and $D$ lie on a circle. Angles $\angle ABD$ and $\angle ACD$ both stand on chord $AD$. Prove that they are equal. 1. Draw the radii $OA$ and $OD$. 2. The angle at the centre standing on chord $AD$ is $\angle AOD$. 3. By the angle at the centre theorem, $\angle AOD = 2\angle ABD$. 4. By the same theorem, $\angle AOD = 2\angle ACD$. 5. Since both angles are half of the same central angle, $\angle ABD = \angle ACD$. ## 5. Opposite angles in a cyclic quadrilateral A **cyclic quadrilateral** is a four-sided shape whose four vertices all lie on the circumference of a circle. The theorem says: opposite angles in a cyclic quadrilateral add to 180°. A quadrilateral $ABCD$ is drawn inside a circle, with its vertices on the circumference. Prove that $\angle ABC + \angle ADC = 180^\circ$. 1. Join $OA$ and $OC$. 2. Angle $\angle ABC$ stands on the arc from $A$ to $C$ that does not contain $B$. 3. Angle $\angle ADC$ stands on the other arc from $A$ to $C$. 4. The two central angles standing on these two arcs add to 360° because they make a full turn around $O$. 5. Each circumference angle is half its matching central angle. 6. Therefore $\angle ABC + \angle ADC$ is half of 360°, so $\angle ABC + \angle ADC = 180^\circ$. In a cyclic quadrilateral, opposite angles add to 180°. Adjacent angles do not necessarily add to 180°. ## 6. The tangent and chord theorem The tangent-chord theorem is often called the **alternate segment theorem**. It says: the angle between a tangent and a chord equals the angle in the opposite segment. You also need the fact that a radius meets a tangent at 90°. A tangent touches a circle at $C$. Points $A$ and $B$ are also on the circumference, and chord $BC$ is drawn. Prove that the angle between the tangent and $BC$ equals $\angle BAC$. 1. Join $OC$ and $OB$. 2. The radius $OC$ is perpendicular to the tangent, so the angle between $OC$ and the tangent is 90°. 3. Let $\angle OCB = y$. Then the angle between the tangent and chord $BC$ is $90^\circ - y$. 4.