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Let's break down the problem step by step:

### Given Information:
- \( TA \) is a tangent to the circle at point \( A \).
- \( \angle BAD = 88^\circ \)
- \( \angle ABD = 46^\circ \)
- \( BC = CD \)

The problem asks:

**(a)** Prove that \( AC \) is the diameter of the circle.

**(b)** Find \( \angle DAT \).

---

### (a) Proving \( AC \) is the diameter of the circle:

#### Step 1: Using the tangent property at \( A \)

Since \( TA \) is a tangent to the circle at point \( A \), the angle between the tangent and the chord \( AD \), which is \( \angle DAT \), must be equal to the angle subtended by the chord \( AD \) on the opposite side of the circle. In other words:
\[
\angle DAT = \angle ACD
\]

#### Step 2: Using the cyclic quadrilateral property

\( ABCD \) forms a cyclic quadrilateral (a quadrilateral inscribed in a circle). In a cyclic quadrilateral, opposite angles sum up to \( 180^\circ \). Thus:
\[
\angle BAD + \angle BCD = 180^\circ
\]
Substituting the given value for \( \angle BAD = 88^\circ \), we get:
\[
88^\circ + \angle BCD = 180^\circ
\]
\[
\angle BCD = 92^\circ
\]

#### Step 3: Using the isosceles triangle property

Since \( BC = CD \), triangle \( BCD \) is isosceles. Therefore, the base angles \( \angle BDC \) and \( \angle BCD \) are equal. Let these angles be \( x \). Thus:
\[
2x + 92^\circ = 180^\circ
\]
Solving for \( x \):
\[
2x = 88^\circ \quad \Rightarrow \quad x = 44^\circ
\]
So, \( \angle BCD = 44^\circ \) and \( \angle BDC = 44^\circ \).

#### Step 4: Angle subtended by the diameter

In a circle, the angle subtended by the diameter at the circumference is always \( 90^\circ \). Since \( \angle ACD = 90^\circ \), it proves that \( AC \) must be the diameter of the circle.

---

### (b) Finding \( \angle DAT \)

From the tangent-secant theorem, we know that:
\[
\angle DAT = \angle ACD
\]
From the cyclic quadrilateral property and previous calculations, we already found that \( \angle ACD = 44^\circ \).

Thus, \( \angle DAT = 44^\circ \).

---

Would you like more details or further explanation? Here are some follow-up questions to expand your understanding:

1. Why is the tangent-secant theorem important in this context?
2. How does the property of cyclic quadrilaterals help in angle calculations?
3. What would change if the circle were not cyclic?
4. How can we verify that \( AC \) is indeed the diameter geometrically?
5. What would be the next step if the triangle were not isosceles?

**Tip**: Always verify relationships between angles in cyclic quadrilaterals using the sum of opposite angles property to avoid calculation errors.

TA is a tangent to the circle ABCD at the point A. It is given that angle BAD = 88°, angle ABD = 46°, and BC = CD. Prove that AC is the diameter of the circle. Also, find the angle DAT.

In this geometry problem, we prove that AC is the diameter of a circle and calculate the angle DAT using properties of cyclic quadrilaterals, the tangent-secant theorem, and isosceles triangles. This problem is suitable for students in Grades 10-12.

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