The problem asks to work out the size of angle $\angle EBA$, where $B$, $C$, $D$, and $E$ are points on the circumference of a circle. We are given that:

• $ED = DB$,
• $DC = CB$, and
• $\angle EDB = 44^\circ$.

Let’s go step by step to solve this problem:

1. Isosceles Triangle $EDB$: Since $ED = DB$, triangle $EDB$ is isosceles. Therefore, the base angles of this triangle are equal. Let’s denote the two equal base angles as $\theta$.

   Since the sum of angles in a triangle is $180^\circ$, we have: $2\theta + 44^\circ = 180^\circ$ Solving for $\theta$: $2\theta = 180^\circ - 44^\circ = 136^\circ$ $\theta = \frac{136^\circ}{2} = 68^\circ$

   So, $\angle BED = \angle BDE = 68^\circ$.

2. Isosceles Triangle $DBC$: Similarly, since $DC = CB$, triangle $DBC$ is also isosceles. Let’s denote the base angles as $\alpha$.

   The angle at $D$ is given as $\angle BDC = \angle BDE - \angle EDB = 68^\circ - 44^\circ = 24^\circ$.

   Now using the triangle angle sum property: $2\alpha + 24^\circ = 180^\circ$ Solving for $\alpha$: $2\alpha = 180^\circ - 24^\circ = 156^\circ$ $\alpha = \frac{156^\circ}{2} = 78^\circ$

   So, $\angle DBC = \angle BDC = 78^\circ$.

3. Cyclic Quadrilateral $ABCE$: Since $B$, $C$, $D$, and $E$ are on the circumference, quadrilateral $ABCE$ is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral sum to $180^\circ$. Hence, we have: $\angle EBA + \angle ECA = 180^\circ$ We already know that $\angle ECA = 78^\circ$ (from step 2), so: $\angle EBA + 78^\circ = 180^\circ$ Solving for $\angle EBA$: $\angle EBA = 180^\circ - 78^\circ = 102^\circ$

Thus, the size of $\angle EBA$ is 102°.

Would you like more details on any of the steps? Here are five related questions to deepen understanding:

1. Why is a triangle with two equal sides called isosceles, and how does it affect the angles?
2. How do we know that the sum of the angles in any triangle is always 180°?
3. Can the properties of cyclic quadrilaterals be applied to any other geometric shape?
4. What other properties do isosceles triangles have, apart from equal angles opposite the equal sides?
5. How does the Circle Theorem help solve problems involving angles in circles?

Tip: Always look out for isosceles triangles in geometry problems—they often help simplify angle calculations!