https://file.aiquickdraw.com/mathbot/file/1728466185972o67nttm2.jpg

Let's walk through the problems from the image one by one.

### Problem 1
In the circle \( ABCD \):
- It is given that \( CA = CB \) and \( \angle ADB = 70^\circ \).
#### (a) Find \( \angle BDC \).

The key insight here is that triangle \( ADB \) is isosceles since \( CA = CB \). Also, note that \( \angle ADB \) is the angle subtended by arc \( AB \), and we can apply the properties of cyclic quadrilaterals and isosceles triangles.

### Problem 2
In the circle \( ABCDE \):
- It is given that \( AB \parallel EC \), \( \angle DAE = 70^\circ \), and \( \angle CED = 30^\circ \).
- \( AD \) and \( CE \) intersect at \( F \).

You need to find expressions for \( m \) and \( k \) in terms of \( \theta \). Here, you can apply the properties of alternate angles, and the intersecting chords theorem will help express the angles in terms of \( \theta \).

### Problem 3
In the circle \( ABCDE \):
- \( BE \) is the diameter of the circle, and it is given that \( AB = CD \) and \( BC : CD = 1:3 \).
- Let \( \angle BDC = \alpha \).

#### (a) Express \( \angle CBD \), \( \angle BCD \), and \( \angle ABD \) in terms of \( \alpha \).
The key step involves using the fact that \( BE \) is the diameter, which means \( \angle BAE = 90^\circ \). You can then express the required angles by using the angle subtended by the same arc property.

#### (b)(i) Express \( \angle ABE \) in terms of \( \alpha \).
Here, you'll apply the cyclic property and the given ratios to find the expression.

#### (b)(ii) It is given that \( G \) is the in-centre of \( \triangle ABD \), and Danny claims that \( G \) lies on \( BE \).
You need to determine whether \( G \), the in-centre, lies on the diameter \( BE \). This can be checked by investigating whether the in-centre (which is where angle bisectors meet) is aligned with the properties of the diameter.

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Would you like to go into the details of any specific part of these problems? Let me know if you want a full solution for one of them, or if you'd like further clarification!

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Here are 5 questions based on this:
1. What is the significance of the property \( CA = CB \) in \( \triangle ADB \)?
2. How do cyclic quadrilateral properties help in solving \( \angle BDC \)?
3. What theorem can be applied to find \( m \) and \( k \) in Problem 2?
4. How do we use the given ratio \( BC : CD = 1:3 \) in Problem 3?
5. Why is the in-centre of a triangle significant for proving Danny's claim in Problem 3(b)(ii)?

**Tip:** Always start by identifying key geometric properties like parallel lines or diameters in a circle, as these often unlock the solution through theorems like alternate angles or the angle subtended by the diameter.

In the figure, ABCD is a circle. It is given that CA = CB and ∠ADB = 70°. Find ∠BDC.

Learn how to solve a geometry problem involving circle theorems, isosceles triangles, and cyclic quadrilaterals. In this problem, you are given a circle with CA = CB and ∠ADB = 70°. Find ∠BDC using properties of cyclic quadrilaterals and isosceles triangles.

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