O is the centre of the circle ABCDE with DEC = 26°, AB=DC and EOC = 106°. Calculate the size of: 6.1 BĈA, 6.2 Â1, 6.3 ÔCD

We are given that $\text{AB} = \text{DC}$. Chords of equal length subtend equal angles at the circumference.
The angle subtended by chord DC at the circumference is $\text{D}\hat{\text{E}}\text{C} = 26^\circ$.
The angle subtended by chord AB at the circumference is $\text{B}\hat{\text{C}}\text{A}$.
Therefore, these angles are equal.

$$\text{B}\hat{\text{C}}\text{A} = \text{D}\hat{\text{E}}\text{C}$$ $$\text{B}\hat{\text{C}}\text{A} = 26^\circ$$

The size of $\text{B}\hat{\text{C}}\text{A}$ is $\boxed{\text{26}^\circ}$.

$\hat{\text{A}}_1$ is the angle $\text{C}\hat{\text{A}}\text{B}$. This is an angle subtended by arc BC at the circumference.
The angle subtended by the same arc BC at the center is $\text{B}\hat{\text{O}}\text{C}$.
We are given $\text{B}\hat{\text{O}}\text{C} = 106^\circ$.
The angle at the center is twice the angle at the circumference subtended by the same arc.

$$\hat{\text{A}}_1 = \frac{1}{2} \times \text{B}\hat{\text{O}}\text{C}$$

$$\hat{\text{A}}_1 = \frac{1}{2} \times 106^\circ$$ $$\hat{\text{A}}_1 = 53^\circ$$

The size of $\hat{\text{A}}_1$ is $\boxed{\text{53}^\circ}$.

The angle subtended by arc DC at the circumference is $\text{D}\hat{\text{E}}\text{C} = 26^\circ$.
The angle at the center $\text{D}\hat{\text{O}}\text{C}$ is twice the angle at the circumference.

$$\text{D}\hat{\text{O}}\text{C} = 2 \times \text{D}\hat{\text{E}}\text{C}$$ $$\text{D}\hat{\text{O}}\text{C} = 2 \times 26^\circ$$ $$\text{D}\hat{\text{O}}\text{C} = 52^\circ$$

Consider $\triangle \text{OCD}$. $\text{OC}$ and $\text{OD}$ are both radii of the circle, so $\text{OC} = \text{OD}$.
Therefore, $\triangle \text{OCD}$ is an isosceles triangle, and its base angles are equal: $\text{O}\hat{\text{C}}\text{D} = \text{O}\hat{\text{D}}\text{C}$.
The sum of angles in a triangle is $180^\circ$.

$$\text{O}\hat{\text{C}}\text{D} + \text{O}\hat{\text{D}}\text{C} + \text{D}\hat{\text{O}}\text{C} = 180^\circ$$

$$\text{O}\hat{\text{C}}\text{D} + \text{O}\hat{\text{C}}\text{D} + 52^\circ = 180^\circ$$ $$2 \times \text{O}\hat{\text{C}}\text{D} = 180^\circ - 52^\circ$$ $$2 \times \text{O}\hat{\text{C}}\text{D} = 128^\circ$$ $$\text{O}\hat{\text{C}}\text{D} = \frac{128^\circ}{2}$$ $$\text{O}\hat{\text{C}}\text{D} = 64^\circ$$

The size of $\text{O}\hat{\text{C}}\text{D}$ is $\boxed{\text{64}^\circ}$.