Question 11

We will use properties of parallel lines and transversals to find the unknown angles.

Step 1 — Find angles at point E using corresponding angles

The line IA and the line GD are parallel. The line HC is a transversal. Angle $\angle ABC$ is given as $\mathbf{45^\circ}$. Angle $\angle KBE$ is vertically opposite to $\angle ABC$. So, $\angle KBE$ is also $\mathbf{45^\circ}$. Angle $\angle KBE$ and $\angle GEH$ are corresponding angles. Corresponding angles are equal when lines are parallel.

$\angle GEH = \angle KBE$

$= 45^\circ$

$\boxed{\angle GEH = 45^\circ}$

The line JF is another transversal. Angle $\angle IKJ$ is given as $\mathbf{78^\circ}$. Angle $\angle BKE$ is vertically opposite to $\angle IKJ$. So, $\angle BKE$ is also $\mathbf{78^\circ}$. Angle $\angle BKE$ and $\angle FED$ are corresponding angles. Corresponding angles are equal when lines are parallel.

$\angle FED = \angle BKE$

$= 78^\circ$

$\boxed{\angle FED = 78^\circ}$

Step 2 — Find angle HEF using angles on a straight line

Angles $\angle GEH$, $\angle HEF$, and $\angle FED$ are on the straight line GD. Angles on a straight line add up to $\mathbf{180^\circ}$. We know $\angle GEH = \mathbf{45^\circ}$ and $\angle FED = \mathbf{78^\circ}$. Let us add these two angles first.

$45^\circ + 78^\circ = 123^\circ$

Now, we subtract this sum from $\mathbf{180^\circ}$ to find $\angle HEF$.

$\angle HEF = 180^\circ - (45^\circ + 78^\circ)$

$= 180^\circ - 123^\circ$

$= 57^\circ$

$\boxed{\angle HEF = 57^\circ}$

Answer

(i) $\angle GEH = \mathbf{45^\circ}$ (ii) $\angle HEF = \mathbf{57^\circ}$ (iii) $\angle FED = \mathbf{78^\circ}$