Some pairs of creases meet inside the paper itself. Others don’t meet within the sheet — but if you extend them far enough in both directions, most of these pairs do eventually meet at a point, because they are not going in exactly the same direction. Only creases that run in exactly the same direction (equally spaced, same slope) will never meet, however far extended — these are parallel lines.

When line l intersects line m, exactly 4 angles are formed at the point of intersection — labelled ∠a, ∠b, ∠c and ∠d.

No. Two distinct straight lines can cross at exactly one point at most. If two “lines” met at two different points, they would have to be the very same line.

1. ∠a and ∠b lie on the straight line m-to-l, so together they form a straight angle: $\angle a + \angle b = 180^\circ$.
2. $\angle b = 180^\circ – 120^\circ = 60^\circ$.
3. ∠b and ∠c form a straight angle too: $\angle c = 180^\circ – 60^\circ = 120^\circ$.
4. ∠c and ∠d form a straight angle: $\angle d = 180^\circ – 120^\circ = 60^\circ$.

1. Since ∠a and ∠b lie on a straight line: $\angle a + \angle b = 180^\circ$.
2. Since ∠a and ∠d also lie on a straight line: $\angle a + \angle d = 180^\circ$.
3. Comparing the two: $\angle a + \angle b = \angle a + \angle d \Rightarrow \angle b = \angle d$.
4. Similarly, $\angle b + \angle a = 180^\circ$ and $\angle b + \angle c = 180^\circ$, so $\angle a = \angle c$.

1. Around a point of intersection, the four angles always add up to $360^\circ$: $\angle a+\angle b+\angle c+\angle d = 360^\circ$.
2. If all four angles are equal, each one is $360^\circ \div 4 = 90^\circ$.
3. This also matches the linear-pair rule: if a linear pair is made of two equal angles, each must be $180^\circ \div 2 = 90^\circ$.

1. FG and FH meet at the endpoint F, at an angle of 115.3° (given).
2. AB and CD intersect each other at an interior point X, forming four angles.
3. ST and UV do not meet within the drawn segments — since S,T,U,V are not evenly spaced/parallel, if extended they would eventually meet at a point.
4. OP and QR also do not meet as drawn; whether they meet on extension depends on whether they are parallel — from the figure they are not perfectly parallel, so extending them would make them meet at a point far away.
5. IL and MJ intersect each other at an interior point Y.

1. The opposite edges of the sheet are parallel to each other.
2. The adjacent edges are perpendicular to each other — they meet at a point and form right angles.
3. After one horizontal fold, a new crease appears midway — you now see 3 parallel lines (2 original horizontal edges + 1 new crease), and this new line is perpendicular to the vertical sides.
4. One more horizontal fold gives 5 parallel lines in total.
5. Folding again follows the pattern $2^n+1$ parallel lines after $n$ horizontal folds — so a third fold gives $2^3+1=9$ parallel lines.
6. A vertical fold creates a new line that is perpendicular to all the horizontal fold-lines.
7. Yes — folding along the diagonal, and then folding the paper again along a line through the same corner, can produce a crease parallel to the original diagonal.

Yes. The two triangular flaps are folded onto the same centre crease from diagonally opposite corners of the square, so corresponding sides of the two triangles are formed by identical folding moves rotated by 180°. That makes edge a parallel to edge p, b parallel to q, and c parallel to r — each pair is related by the symmetry of the square about its centre.

A line is perpendicular to a given dot-paper line when it meets it at a right angle. On square dot paper, this is easiest to check for horizontal/vertical lines (rotate 90°) and for 45°-diagonal lines (the perpendicular diagonal runs the opposite way). Try drawing one perpendicular line for each segment shown, meeting it at a right angle at a dot.

1. (a) Perpendicular lines were spotted where two edges of a shape met at a 90° corner — visually, a “square” corner rather than a slanted one.
2. (b) Parallel lines were spotted by checking that two segments stayed the same distance apart along their whole length, and had the same slope on the dot grid (moved the same number of dots across for the same number of dots up).

This is open-ended — any two (or more) segments that keep the exact same slope and constant spacing count as a parallel family. E.g. two segments that each move “3 dots across, 1 dot up” from their starting dot are automatically parallel to each other, no matter where they start.

1. (a) Yes, a little — segments at “odd” slopes (not horizontal, vertical, or 45°) are harder to copy exactly.
2. (b) The trickiest ones are the segments with uneven slopes, such as those labelled e, f, h and g, since their rise-over-run isn’t a simple 1:1 or 1:0 ratio.
3. (c) By counting how many dots the line moved across and how many dots it moved up/down, then repeating that exact same horizontal and vertical step from a different starting dot.

Line c is parallel to line a. Tracing along a and c, both keep the exact same steady slant and stay the same distance apart everywhere — while b tilts at a visibly different, steeper angle and would eventually meet a if extended, so it isn’t parallel.

1. No — vertically opposite angles are always equal: $\angle1=\angle3$, $\angle2=\angle4$, $\angle5=\angle7$, $\angle6=\angle8$. So all eight cannot be different.
2. Even asking for just five different values among ∠2, ∠3, ∠4, ∠5, ∠6 fails: at the top intersection, $\angle2=\angle4$ (vertically opposite), so at most only 4 distinct measures can ever appear among all eight angles.

1. Draw a line l and a transversal t, meeting at X. Measure one angle, say $\angle a = 60^\circ$ (the linear pair partner is automatically $120^\circ$) — two distinct measures so far.
2. Mark a point Y further along t, and draw a second line m through Y that makes the same 60° angle with t (copy ∠a using tracing paper or a protractor).
3. Since ∠a = ∠b = $60^\circ$ and both linear-pair partners are $120^\circ$, only two distinct angle values appear anywhere in the figure — and lines l and m turn out to be parallel.

Yes — the traced copy of ∠a fits exactly over ∠b, and every other corresponding pair matches too. This confirms the rule: corresponding angles formed by a transversal on a pair of parallel lines are always equal.

It is tricky to do just by eye — the transversal has to be drawn so that it strikes both lines at the exact same angle. In practice this only happens reliably when l and m are already parallel; if they are not parallel, no transversal can ever make the corresponding angles equal (see the rule below).

1. Both new lines were drawn perpendicular to line l using the set square, so each makes a $90^\circ$ angle with l.
2. Treating l as a transversal to the two new lines, the two $90^\circ$ angles are corresponding angles — and they are equal.
3. Since corresponding angles are equal, the two new lines must be parallel (this is guaranteed, not just “looks parallel”).

1. Place a set square so one edge lies exactly along line l.
2. Hold a ruler firmly against another edge of the set square (to act as a guide rail).
3. Slide the set square along the ruler, keeping the ruler fixed, until the sliding edge reaches point A.
4. Draw a line along that edge of the set square, through A.

Both l and m are perpendicular to the same crease t, so treating t as a transversal, both intersections make $90^\circ$ angles with it — equal corresponding angles. Equal corresponding angles on a transversal always mean the two lines are parallel, so $l \parallel m$.

1. ∠b is the corresponding angle of ∠f (same position at the two crossings), and since $l\parallel m$: $\angle b=\angle f=120^\circ$.
2. ∠d is vertically opposite to ∠b, so $\angle d=\angle b=120^\circ$.
3. Hence $\angle f=\angle b=\angle d$, i.e. $\angle f=\angle d$ — true for any measure of ∠f, not just $120^\circ$.

1. ∠2 is the corresponding angle of ∠6, and $l\parallel m$, so $\angle2=135^\circ$.
2. ∠8 is vertically opposite to ∠6: $\angle8=135^\circ$. ∠4 is the corresponding angle of ∠8: $\angle4=135^\circ$.
3. So $\angle2=\angle4=\angle6=\angle8=135^\circ$.
4. ∠5 and ∠6 form a linear pair: $\angle5=180^\circ-135^\circ=45^\circ$.
5. By the same pattern, $\angle1=\angle3=\angle7=45^\circ$.

1. ∠a and ∠b form a linear pair: $\angle b=180^\circ-120^\circ=60^\circ$.
2. ∠b is the corresponding angle of ∠f. If $l\parallel m$, we would need $\angle b=\angle f$.
3. But $\angle b=60^\circ$ while $\angle f=70^\circ$ — they are not equal.

1. ∠2 and ∠3 form a linear pair: $\angle2=180^\circ-50^\circ=130^\circ$.
2. ∠2 and ∠6 are corresponding angles, and $l\parallel m$, so $\angle6=\angle2$.

1. AD is a transversal of the parallel lines AB and CD, so interior angles on the same side add to $180^\circ$: $\angle ADC+\angle DAB=180^\circ \Rightarrow \angle DAB = 180^\circ-60^\circ=120^\circ$.
2. $\angle DAB=\angle DAC+\angle CAB \Rightarrow 120^\circ = 65^\circ+\angle CAB \Rightarrow \angle CAB = 55^\circ$.
3. Now CD is a transversal of parallel segments AD and BC: $\angle ADC+\angle BCD=180^\circ \Rightarrow \angle BCD=180^\circ-60^\circ=120^\circ$.
4. Similarly, using AB as transversal of AD and BC, $\angle ABC = 60^\circ$.

1. ∠a and the 48° angle are vertically opposite angles (formed by the same pair of crossing lines).
2. Vertically opposite angles are always equal.

1. b° and 52° are corresponding angles at matching positions on the two parallel lines.
2. Corresponding angles on parallel lines cut by a transversal are always equal.

1. 99° and 81° are a linear pair at the top intersection: $99^\circ+81^\circ=180^\circ$ ✓
2. c° is the corresponding angle of the 81° angle, so c° = 81° since the lines are parallel.

1. 81° and 99° form a linear pair: $81^\circ+99^\circ=180^\circ$ ✓
2. d° is the corresponding angle of the 99° angle (matching position at the other parallel line), so d° = 99°.

1. Check: 97° and 83° form a linear pair ($97^\circ+83^\circ=180^\circ$), confirming the lines drawn are straight.
2. e° is the alternate angle of the 69° angle across the transversal, and since the lines are parallel, alternate angles are equal.

1. 132° and f° are interior angles on the same side of the transversal (co-interior / allied angles).
2. Co-interior angles on parallel lines always add up to $180^\circ$: $f = 180^\circ – 132^\circ = 48^\circ$.

1. 58° and 122° form a linear pair at their intersection ($58^\circ+122^\circ=180^\circ$).
2. g° corresponds to the 122° angle at the matching position on the other parallel line, so g° = 122°.

1. h° is formed at the point where the slanted line crosses the upper parallel line.
2. h° and the marked 75° angle are corresponding/alternate angles across the two parallel lines, so h° = 75°.

1. i° is the alternate angle of the marked 54° angle, formed across the transversal between the two slanted lines (which act as the pair of lines here).
2. Since the relevant pair of lines are parallel by construction, the alternate angles are equal: i° = 54°.

1. The three slanted lines are parallel, so the transversal makes equal corresponding angles with each of them.
2. j° corresponds to the 97° angle marked at the first crossing, so j° = 97°.

1. a° and the 42° angle sit next to each other on the same straight line, so they form a linear pair: $a+42^\circ=180^\circ$.
2. $a = 180^\circ-42^\circ=138^\circ$.

1. 62° and a° are co-interior (allied) angles on the same side of the transversal, between the parallel lines.
2. Co-interior angles sum to $180^\circ$: $a = 180^\circ – 62^\circ = 118^\circ$.

1. Draw an auxiliary line through the middle bend-point, parallel to all three lines.
2. The 110° angle at the top is co-interior with part of the angle at the bend: that part $=180^\circ-110^\circ=70^\circ$.
3. This splits the bend so the remaining transversal segment carries the other 35° forward as an alternate angle down to the bottom line.
4. Adding the two parts at the bottom: $a = 70^\circ+35^\circ = 105^\circ$.

1. The right-angle mark shows the base angle is $90^\circ$, split into the 67° part and an adjacent part.
2. The adjacent part $=90^\circ-67^\circ=23^\circ$, and a° is alternate/corresponding to this part across the parallel lines.
3. $a = 23^\circ$.

1. The vertical and horizontal lines meet at a right angle: $90^\circ$, split into $65^\circ$ and x°: $x = 90^\circ – 65^\circ = 25^\circ$.
2. x° and y° lie together on the straight slanted line as a linear pair with the angle on the other side: $y = 180^\circ – x = 180^\circ – 25^\circ = 155^\circ$.

1. x° is the exterior angle of the small triangle formed by the two transversals and the base line.
2. By the exterior-angle idea (equivalent to subtracting the base angles measured from the same line): $x = 78^\circ – 53^\circ = 25^\circ$.

1. GD ∥ IA, with the line through B and E as one transversal: ∠GEH and ∠ABC are corresponding angles, so $\angle GEH = \angle ABC = 45^\circ$.
2. The line through K and E is a second transversal: ∠FED and ∠IKJ are corresponding angles, so $\angle FED = \angle IKJ = 78^\circ$.
3. G, E, D lie on the same straight line, so the three angles at E on one side add to $180^\circ$: $\angle GEH+\angle HEF+\angle FED = 180^\circ$.
4. $\angle HEF = 180^\circ – 45^\circ – 78^\circ = 57^\circ$.

1. CD and EF are parallel vertical lines, cut by the transversal BEF: ∠BEF and y° are co-interior (same-side interior) angles, so $y = 180^\circ – 55^\circ = 125^\circ$.
2. AB ∥ CD, cut by the same transversal: x° and y° are corresponding angles, so $x = y = 125^\circ$.

1. Draw an auxiliary line through N parallel to LM. Since LM ∥ this auxiliary line, alternate angle with ∠LMN ($40^\circ$) gives the lower split of ∠MNO as $40^\circ$; so the remaining part (between the auxiliary line and NO) is $96^\circ – 40^\circ = 56^\circ$.
2. Draw a second auxiliary line through O, parallel to the first (hence also parallel to LM and PQ). By alternate angles, the angle between ON and this new line equals the $56^\circ$ found above.
3. Similarly, the angle between this line and OP is alternate to ∠OPQ’s $52^\circ$, so it equals $52^\circ$.
4. $\angle NOP$ is the sum of these two parts: $\angle NOP = 56^\circ + 52^\circ = 108^\circ$.