Question

a, b, c and d are points on a circle.
pcq is a tangent to the circle.
ab = cb.
angle bcq = x°
prove that angle cda = 2x°
give reasons for each stage in your working.
diagram not accurately drawn

Explanation:

Step1: Use Alternate Segment Theorem

The Alternate Segment Theorem states that the angle between the tangent and a chord is equal to the angle in the alternate segment. So, $\angle BCQ = \angle BAC = x^{\circ}$ (Alternate Segment Theorem, since $PCQ$ is tangent and $CB$ is a chord).

Step2: Use Isosceles Triangle Property

Given $AB = CB$, triangle $ABC$ is isosceles with $\angle BAC=\angle BCA = x^{\circ}$. Then, by the angle - sum property of a triangle ($\angle BAC+\angle BCA+\angle ABC = 180^{\circ}$), we can find $\angle ABC=180 - x - x=180 - 2x^{\circ}$. But we can also use the property of cyclic quadrilaterals or the fact that the angle subtended by an arc at the center is twice the angle subtended at the circumference. Alternatively, we can use the property that the angle in the same segment: $\angle CDA$ and $\angle CBA$ are related. Wait, actually, the angle subtended by arc $CA$ at the circumference: since $AB = CB$, arc $AB$ = arc $CB$. The angle $\angle CDA$ subtends arc $CBA$, and $\angle BAC$ subtends arc $BC$. Wait, let's re - approach.

Since $AB = CB$, the arcs $AB$ and $CB$ are equal. The angle $\angle BAC$ (which is $x^{\circ}$) subtends arc $BC$, so the angle subtended by arc $AB$ (which is equal to arc $BC$) at the circumference is also $x^{\circ}$. Now, the angle $\angle CDA$ subtends arc $CBA$ (arc $CB +$ arc $BA$). Since arc $CB$ and arc $BA$ each subtend an angle of $x^{\circ}$ at the circumference, the angle subtended by arc $CBA$ at the circumference (angle $\angle CDA$) is $x^{\circ}+x^{\circ}=2x^{\circ}$.

Wait, a better way:

1. By Alternate Segment Theorem: $\angle BCQ=\angle BAC = x^{\circ}$ (angle between tangent $PCQ$ and chord $CB$ is equal to the angle in the alternate segment, i.e., $\angle BAC$).
2. Since $AB = CB$, $\triangle ABC$ is isosceles with $\angle BAC=\angle BCA=x^{\circ}$.
3. The angle $\angle ABC = 180^{\circ}-\angle BAC-\angle BCA=180 - 2x^{\circ}$. But $ABCD$ is a cyclic quadrilateral? Wait, no, $A,B,C,D$ are on a circle, so $ABCD$ is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of opposite angles is $180^{\circ}$, but also, the angle subtended by an arc at the circumference: the angle $\angle CDA$ and $\angle CBA$: Wait, actually, the angle $\angle CDA$ and $\angle CBA$: no, the angle $\angle CDA$ subtends arc $CA$, and $\angle CBA$ subtends arc $CA$? Wait, no. Let's use the property that the angle between tangent and chord is equal to the angle in the alternate segment, and then use the isosceles triangle and the fact that the angle at the center is twice the angle at the circumference.

Wait, let's start over:

- Step 1: Alternate Segment Theorem: $\angle BCQ=\angle BAC=x^{\circ}$ (tangent $PCQ$, chord $CB$, so angle between tangent and chord $CB$ is equal to the angle in the alternate segment, which is $\angle BAC$).
- Step 2: $AB = CB$, so $\triangle ABC$ is isosceles with $\angle BAC=\angle BCA = x^{\circ}$.
- Step 3: The angle $\angle ABC=180^{\circ}-\angle BAC - \angle BCA=180 - 2x^{\circ}$. But in the cyclic quadrilateral $ABCD$, $\angle CDA+\angle ABC = 180^{\circ}$? No, that's not right. Wait, no, the sum of opposite angles in a cyclic quadrilateral is $180^{\circ}$, but $\angle CDA$ and $\angle CBA$: Wait, maybe I made a mistake. Let's use the property that the angle subtended by an arc at the center is twice the angle subtended at the circumference. The arc $CB$ subtends $\angle BAC=x^{\circ}$ at the circumference, so the arc $CB$ subtends $2x^{\circ}$ at the center. Since $AB = CB$, arc $AB$ is equal to arc $CB$, so arc $AB$ also subtends $2x^{\circ}$ at the center. Then arc $CBA$ (arc $CB +$ arc $AB$) subtends $2x^{\circ}+2x^{\circ}=4x^{\circ}$ at the center? No, that's not correct. Wait, the angle subtended by arc $CA$ at the center: arc $CA$ is arc $CB +$ arc $BA$. Since arc $CB$ and arc $BA$ are equal (because $AB = CB$), let the measure of arc $CB$ be $m$, then arc $BA$ is also $m$. The angle $\angle BAC$ subtends arc $CB$, so $m = 2x^{\circ}$ (since the angle subtended at the center is twice the angle subtended at the circumference). Then arc $CA$ has measure $m + m=2m = 4x^{\circ}$? No, the angle subtended by arc $CB$ at the circumference is $\angle BAC=x^{\circ}$, so the measure of arc $CB$ is $2x^{\circ}$ (because angle subtended at center is $2\times$ angle at circumference). Since $AB = CB$, arc $AB$ is also $2x^{\circ}$. Then the angle subtended by arc $CA$ (arc $CB +$ arc $AB$) at the circumference is $\angle CDA$. The measure of arc $CA$ is $2x^{\circ}+2x^{\circ}=4x^{\circ}$, so the angle subtended at the circumference is $\frac{1}{2}\times4x^{\circ}=2x^{\circ}$. Yes, that works.

So:

1. By Alternate Segment Theorem: $\angle BCQ=\angle BAC = x^{\circ}$ (angle between tangent $PCQ$ and chord $CB$ is equal to the angle in the alternate segment $\angle BAC$).
2. The angle subtended by an arc at the center is twice the angle subtended at the circumference. So, the arc $CB$ subtends $\angle BAC$ at the circumference, so the measure of arc $CB$ is $2\angle BAC = 2x^{\circ}$.
3. Since $AB = CB$, the lengths of chords $AB$ and $CB$ are equal, so the measures of arcs $AB$ and $CB$ are equal. Thus, the measure of arc $AB$ is also $2x^{\circ}$.
4. The arc $CA$ is composed of arc $CB$ and arc $AB$, so the measure of arc $CA$ is $2x^{\circ}+2x^{\circ}=4x^{\circ}$.
5. The angle $\angle CDA$ subtends arc $CA$ at the circumference. By the property that the angle subtended by an arc at the circumference is half the angle subtended at the center, $\angle CDA=\frac{1}{2}\times$ measure of arc $CA$.
6. Substituting the measure of arc $CA = 4x^{\circ}$, we get $\angle CDA=\frac{1}{2}\times4x^{\circ}=2x^{\circ}$.

Answer:

To prove that $\angle CDA = 2x^{\circ}$, we proceed as follows:

Step 1: Apply the Alternate Segment Theorem

The Alternate Segment Theorem states that the angle between a tangent and a chord is equal to the angle in the alternate segment. For tangent $PCQ$ and chord $CB$, we have:
$\angle BCQ=\angle BAC = x^{\circ}$ (Alternate Segment Theorem).

Step 2: Relate Arc Length to Angle Subtended at Circumference

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circumference. The arc $CB$ subtends $\angle BAC$ (which is $x^{\circ}$) at the circumference. Thus, the measure of arc $CB$ is:
$\text{Measure of arc }CB = 2\times\angle BAC=2x^{\circ}$ (Angle - Subtended - by - Arc property).

Step 3: Use $AB = CB$ to Find Arc $AB$

Since $AB = CB$, the chords $AB$ and $CB$ are equal in length. In a circle, equal - length chords subtend equal - length arcs. Thus, the measure of arc $AB$ is also $2x^{\circ}$ (Equal chords subtend equal arcs).

Step 4: Find the Measure of Arc $CA$

The arc $CA$ is composed of arc $CB$ and arc $AB$. So, the measure of arc $CA$ is:
$\text{Measure of arc }CA=\text{Measure of arc }CB+\text{Measure of arc }AB = 2x^{\circ}+2x^{\circ}=4x^{\circ}$.

Step 5: Find $\angle CDA$ Using Arc - Angle Relationship

The angle $\angle CDA$ subtends arc $CA$ at the circumference. By the angle - subtended - by - arc property (angle at circumference is half the angle at center, or directly, angle subtended by an arc at the circumference is half the measure of the arc), we have:
$\angle CDA=\frac{1}{2}\times\text{Measure of arc }CA$.

Substituting the measure of arc $CA = 4x^{\circ}$, we get:
$\angle CDA=\frac{1}{2}\times4x^{\circ}=2x^{\circ}$.

Thus, we have proven that $\angle CDA = 2x^{\circ}$.