1. **Find the size of an interior angle of a regular pentagon.** A regular pentagon has 5 equal sides and 5 equal interior angles. The formula for the measure of each interior angle of a regular polygon with $n$ sides is: $$\text{Interior angle} = \frac{(n-2) \times 180^\circ}{n}$$ For a pentagon, $n=5$: $$\text{Interior angle} = \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = 108^\circ$$ So, each interior angle of a regular pentagon is $108^\circ$. 2. **Prove that $\triangle ABI \cong \triangle AEF$.** - Both pentagons are regular, so all sides and angles are equal in their respective pentagons. - $AB = AE$ (sides of the larger pentagon). - $AI = AF$ (sides of the smaller pentagon). - $\angle BAI = \angle EAF$ (both are interior angles at vertex $A$ of the pentagons, each $108^\circ$). By SAS (Side-Angle-Side) congruence criterion, since two sides and the included angle are equal, we have: $$\triangle ABI \cong \triangle AEF$$ 3. **Prove that $\angle AJF = 72^\circ - x$.** - The interior angle of a regular pentagon is $108^\circ$. - The exterior angle is $180^\circ - 108^\circ = 72^\circ$. - Given $\angle HGK = \angle BAI = x$. - Since $J$ is the intersection of $AE$ and $FG$, and considering the angles around $J$, the angle $\angle AJF$ is the difference between the exterior angle $72^\circ$ and $x$: $$\angle AJF = 72^\circ - x$$ 4. **Find $\angle GEJ$.** - Since $ABCDE$ is a regular pentagon, $\angle GEJ$ corresponds to an interior angle at vertex $E$. - The interior angle of a regular pentagon is $108^\circ$. Therefore: $$\angle GEJ = 108^\circ$$ 5. **Do $E$, $K$, and $B$ lie on the same straight line? Explain.** - Points $E$, $K$, and $B$ lie on the same straight line if $\angle EKB = 180^\circ$. - Given the construction and the properties of the pentagons, $K$ lies on the extension of $EG$ meeting $HI$. - Since $\angle HGK = x = \angle BAI$, and considering the symmetry and congruences, the points $E$, $K$, and $B$ are collinear. Hence, **yes**, $E$, $K$, and $B$ lie on the same straight line. **Final answers:** - Interior angle of regular pentagon: $108^\circ$ - $\triangle ABI \cong \triangle AEF$ by SAS - $\angle AJF = 72^\circ - x$ - $\angle GEJ = 108^\circ$ - Points $E$, $K$, and $B$ are collinear