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.
∠bcq=∠bac (a.s.t)
=x°
∠acb = x° (isoceles δ)

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, $\angle BCQ = \angle BAC$. Given $\angle BCQ = x^\circ$, then $\angle BAC = x^\circ$.

Step2: Use Isosceles Triangle Property

Since $AB = CB$, $\triangle ABC$ is isosceles with $AB = CB$. In an isosceles triangle, the angles opposite equal sides are equal. So, $\angle BAC = \angle ACB$. We know $\angle BAC = x^\circ$, so $\angle ACB = x^\circ$.

Step3: Calculate $\angle ABC$

The sum of angles in a triangle is $180^\circ$. In $\triangle ABC$, $\angle ABC + \angle BAC + \angle ACB = 180^\circ$. Substituting the known angles: $\angle ABC + x^\circ + x^\circ = 180^\circ$, so $\angle ABC = 180^\circ - 2x^\circ$.

Step4: Use Cyclic Quadrilateral Property

In a cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of opposite angles is $180^\circ$. Also, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Alternatively, the angle subtended by a chord in the alternate segment and the angle in the same segment: $\angle CDA$ and $\angle CBA$ are related? Wait, correction: The angle $\angle CDA$ and $\angle CBA$: Wait, actually, the angle at the center or the angle in the same segment. Wait, another approach: The angle between tangent and chord is equal to the angle in the alternate segment, and also, in cyclic quadrilateral $ABCD$, $\angle CDA + \angle CBA = 180^\circ$? No, wait, no: Wait, $\angle CDA$ and $\angle CBA$: Wait, no, the correct property is that the angle subtended by a chord $BC$ at the circumference: $\angle BAC = \angle BDC$? No, wait, let's re - express.

Wait, we have $\angle BCQ = x^\circ$ (tangent - chord angle), $\angle BAC = x^\circ$ (alternate segment theorem). $AB = CB$, so $\angle BAC=\angle BCA = x^\circ$ (isosceles triangle). Then $\angle ABC = 180 - 2x$. Now, in 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 $180^\circ$, so $\angle CDA + \angle CBA = 180^\circ$? Wait, no, $\angle DAB + \angle DCB = 180^\circ$, $\angle CDA + \angle CBA = 180^\circ$. Wait, but we found $\angle CBA = 180 - 2x$, so $\angle CDA = 180 - (180 - 2x)=2x^\circ$. Yes, that's it. Because in cyclic quadrilateral $ABCD$, $\angle CDA + \angle CBA = 180^\circ$ (opposite angles of cyclic quadrilateral are supplementary). We found $\angle CBA = 180^\circ - 2x^\circ$, so $\angle CDA = 180^\circ-(180^\circ - 2x^\circ)=2x^\circ$.

Let's re - structure the steps properly:

Step1: Alternate Segment Theorem

By the Alternate Segment Theorem, the angle between tangent $PCQ$ and chord $BC$ is equal to the angle in the alternate segment. So, $\angle BCQ=\angle BAC$. Given $\angle BCQ = x^\circ$, so $\angle BAC = x^\circ$.

Step2: Isosceles Triangle ($\triangle ABC$)

Since $AB = CB$, $\triangle ABC$ is isosceles with $AB = CB$. Therefore, $\angle BAC=\angle BCA$ (angles opposite equal sides in an isosceles triangle). Since $\angle BAC = x^\circ$, then $\angle BCA = x^\circ$.

Step3: Sum of Angles in $\triangle ABC$

The sum of the interior angles of a triangle is $180^\circ$. In $\triangle ABC$, we have $\angle ABC+\angle BAC+\angle BCA = 180^\circ$. Substituting $\angle BAC = x^\circ$ and $\angle BCA = x^\circ$, we get $\angle ABC + x^\circ+x^\circ=180^\circ$. Solving for $\angle ABC$, we have $\angle ABC=180^\circ - 2x^\circ$.

Step4: Cyclic Quadrilateral Property (Supplementary Opposite Angles)

$ABCD$ is a cyclic quadrilateral (since $A, B, C, D$ lie on a circle). In a cyclic quadrilateral, the sum of a pair of opposite angles is $180^\circ$. So, $\angle CDA+\angle ABC = 180^\circ$.

Step5: Solve for $\angle CDA$

Substitute $\angle ABC = 180^\circ - 2x^\circ$ into the equation $\angle CDA+\angle ABC = 180^\circ$:

$\angle CDA+(180^\circ - 2x^\circ)=180^\circ$

Subtract $180^\circ - 2x^\circ$ from both sides:

$\angle CDA=180^\circ-(180^\circ - 2x^\circ)=2x^\circ$

Answer:

By using the Alternate Segment Theorem, Isosceles Triangle Property, Sum of Angles in a Triangle, and Cyclic Quadrilateral Property (supplementary opposite angles), we have proven that $\angle CDA = 2x^\circ$.