Question: In the figure, $EF$ is parallel to $HI$. $G$ lies between the two lines, with $\angle EFG = 114°$ and $\angle GHI = 38°$. Find $\angle HGF$.

[diagram in original sheet: two parallel lines with the bent path F–G–H between them]

Step 1. The angle at $G$ doesn't touch either parallel line directly — so draw a helper line $GA$ through $G$, parallel to both $EF$ and $HI$.

This splits $\angle HGF$ into two pieces, one against each parallel line.

Step 2. Upper piece, using $EF$ // $GA$:

$$\angle AGF = 180° - 114° = 66° \quad (\text{int } \angle\text{s})$$

Step 3. Lower piece, using $GA$ // $HI$:

$$\angle HGA = 38° \quad (\text{alt } \angle\text{s})$$

Step 4. Add the two pieces:

$$\angle HGF = 66° + 38° = 104°$$

Check. $\angle HGF$ should be obtuse from the sketch, and $104°$ is. ✓

⚠ Watch out: The helper line must be drawn parallel to the given pair — through the "bend" point. Trying to force alternate or interior angles directly between $F$ and $H$ (skipping the construction) is how this figure gets mis-chased; each property only works across one pair of parallel lines at a time.

Question: In the diagram, $BCD$ is a straight line. $\angle BAC = x°$, $\angle ABC = (x+10)°$ and the exterior angle $\angle ACD = (230-2x)°$. Find the value of $x$, and hence $\angle ACB$.

[diagram in original sheet: triangle ABC with BC extended to D, exterior angle marked at C]

Step 1. The exterior angle of a triangle equals the sum of the two interior opposite angles:

$$x + (x + 10) = 230 - 2x \quad (\text{ext } \angle \text{ of } \triangle)$$

Step 2. Simplify and collect $x$ on one side:

$$2x + 10 = 230 - 2x$$ $$4x = 220$$

Step 3. Solve:

$$x = 55$$

Step 4. Now the angles are $\angle BAC = 55°$ and $\angle ABC = 65°$, so inside the triangle:

$$\angle ACB = 180° - 55° - 65° = 60° \quad (\angle \text{ sum of } \triangle)$$

Check. Exterior angle: $230 - 2(55) = 120°$, and $\angle ACB + 120° = 180°$ ✓ ($\angle$s on a straight line) — both routes agree.

⚠ Watch out: The exterior angle pairs with the two opposite interior angles ($A$ and $B$), never with its neighbour $\angle ACB$ — with $\angle ACB$ it makes a straight line instead. Both facts are true; mixing them up builds the wrong equation.

Question: $ABCD$ is a rhombus whose diagonals $AC$ and $BD$ meet at $O$. Given $\angle CDO = 55°$, find $\angle ADC$ and $\angle DAC$.

[diagram in original sheet: rhombus with both diagonals drawn, O at the crossing]

Step 1. Recall what a rhombus gives you: all four sides equal, opposite angles equal, and each diagonal bisects the corner angles it joins; the diagonals also cross at $90°$.

Step 2. $DO$ lies along the diagonal $DB$, which bisects the corner angle $\angle ADC$:

$$\angle ADC = 2 \times \angle CDO = 2 \times 55° = 110°$$

Step 3. For $\angle DAC$, look at $\triangle ADC$. Its sides $AD$ and $DC$ are sides of the rhombus, so $AD = DC$ — the triangle is isosceles with apex angle $110°$ at $D$:

$$\angle DAC = \angle DCA = \dfrac{180° - 110°}{2} = 35°$$

Check. In $\triangle ODC$: $\angle DOC = 90°$ (diagonals of a rhombus), $\angle CDO = 55°$, so $\angle DCO = 35°$ ✓ — matches $\angle DCA$.

⚠ Watch out: Diagonals bisect the corner angles in a rhombus and a square, but not in a general parallelogram or rectangle — quote the right shape's property. The $90°$ crossing is also rhombus/square-only.