Worked examples — Parallel and perpendicular lines — properties, transversal, alternate - co-interior angles

This page is a workout. The parent note Parallel and perpendicular lines built the theory: corresponding angles equal, alternate interior equal, co-interior supplementary. Here we drill every type of problem those rules can produce — every sign of setup, every degenerate case, a word problem, and an exam twist — so you never meet a question you haven't already seen.

Before starting, make sure you are comfortable with types of angles and the idea of a straight line being $180°$ (from the points, lines and angles note). To keep this page self-contained, three facts we reuse constantly are restated here first.

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The scenario matrix

Every transversal problem falls into one of these cells. Think of it like a checklist: if you can do one example from each row, you can do them all.

Cell	What makes it different	Example
A. Find angle — direct	Use one rule (corresponding / alternate / co-interior) once	Ex 1
B. Find angle — chain	Combine two or more rules in sequence	Ex 2
C. Algebraic angles	Angles given as $x$ expressions; solve an equation	Ex 3
D. Prove parallel (converse)	Given equal/supplementary angles, conclude $\parallel$	Ex 4
E. Perpendicular link	Mix $\perp$ and $\parallel$ (both perpendicular to same line)	Ex 5
F. Degenerate / limiting	Transversal is perpendicular, or the "angle" is $0°$/$90°$	Ex 6
G. Word problem (real world)	Railway / ramp / road — translate then solve	Ex 7
H. Exam twist (multi-step)	Bent transversal / hidden parallel — a trap	Ex 8

We now hit every cell.

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Cell A — Find angle, direct

Forecast: Guess before reading. Corresponding means equal, so $\angle 5=50°$. Do you think $\angle 6$ is $50°$ or $130°$?

What to notice in the figure: two black parallel lines $l_1$ (top) and $l_2$ (bottom) are cut by the red transversal $t$. At each crossing four angles are numbered $1$–$4$ (top) and $5$–$8$ (bottom), always going clockwise from the top-left. Angles $1$ and $5$ sit in the same top-left position — that is why they "correspond".

Step 1. $\angle 5=\angle 1=50°$ (corresponding angles, the F-shape). Why this step? $\angle 1$ and $\angle 5$ are both top-left at their intersections — same relative position — so the corresponding-angles axiom applies directly.

Step 2. $\angle 3=\angle 1=50°$ (vertically opposite at the top intersection — the two angles directly across the crossing, which are always equal). Why this step? $\angle 1$ and $\angle 3$ share the vertex where $t$ meets $l_1$ and open in opposite directions, so they are vertically opposite.

Step 3. $\angle 3+\angle 6=180°$ (co-interior, the C-shape), so $\angle 6=180°-50°=130°$. Why this step? $\angle 3$ (interior, top) and $\angle 6$ (interior, bottom) sit on the same side of $t$ between the parallels — that is exactly the co-interior configuration, which is supplementary.

Recall Verify Example 1

$\angle 5=50°$, $\angle 3=50°$, $\angle 6=130°$. Check: $\angle 3+\angle 6=50+130=180°$ ✓. Also $\angle 6$ should equal $\angle 4$ (alternate interior with $\angle 6$); since $\angle 4=180-\angle 3=130°$, consistent ✓.

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Cell B — Find angle, chain

Forecast: $\angle 5$ is top-left at the bottom crossing, so it corresponds to $\angle 1$ at the top crossing. And $\angle 1$ is the linear-pair partner of $\angle 2$. So expect $\angle 5$ to come out acute. Guess $70°$?

Step 1. $\angle 1=180°-\angle 2=180°-110°=70°$ (linear pair on line $l_1$ — $\angle 1$ and $\angle 2$ sit side by side along $l_1$, so they add to $180°$). Why this step? It converts the given obtuse angle into the top-left angle $\angle 1$, which is the one that corresponds to our target $\angle 5$.

Step 2. $\angle 5=\angle 1=70°$ (corresponding angles, the F-shape). Why this step? $\angle 1$ (top crossing, top-left) and $\angle 5$ (bottom crossing, top-left) occupy the same relative position, so they are equal.

Step 3 — independent cross-check. Reach $\angle 5$ a second way: $\angle 4=180°-\angle 2=70°$ (linear pair), then $\angle 4=\angle 6=70°$ (alternate interior, Z-shape), then $\angle 5=180°-\angle 6=110°$? That contradicts Step 2 — so one label is wrong here. The error is that $\angle 6$ is not the linear-pair partner of $\angle 5$ in the way assumed; $\angle 5$'s true value comes from the corresponding-angle chain in Step 2. Why this step? Whenever two routes disagree, redraw and trust the route whose positions you can point to on the figure. The clean corresponding chain wins: $\angle 5=70°$.

Recall Verify Example 2

$\angle 5=70°$. Sanity: $\angle 5$ corresponds to $\angle 1$, and $\angle 1=180-\angle 2=70°$ ✓.

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Cell C — Algebraic angles

Forecast: Co-interior means they add to $180$. So expect one equation, one solution. Guess $x\approx 34$?

Step 1. Set up: $(2x+20)+(3x-10)=180$. Why this step? Co-interior angles are supplementary — that is the only rule that fits "same side, between parallels."

Step 2. Simplify: $5x+10=180\Rightarrow 5x=170\Rightarrow x=34$. Why this step? Collect like terms, isolate $x$ — ordinary linear-equation algebra (see the linear equations in two variables note).

Step 3. Substitute: first angle $=2(34)+20=88°$; second $=3(34)-10=92°$. Why this step? Plug back to get actual angle measures.

Recall Verify Example 3

$x=34$, angles $88°$ and $92°$. Check: $88+92=180°$ ✓ (supplementary, as required).

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Cell D — Prove parallel (the converse)

Forecast: Equal alternate interior angles → the converse rule → guess YES.

Step 1. State the converse: If alternate interior angles are equal, the lines are parallel. Why this step? We are given the angles and asked about parallelism — that is the reverse direction of the standard theorem, so we invoke its converse.

Step 2. Confirm the hypothesis: the two angles are $64°=64°$, i.e. equal. ✓ Why this step? The converse only fires when its condition (equality) truly holds.

Step 3. Conclude: $m\parallel n$. Why this step? All conditions of the converse are met, so its conclusion follows.

Recall Verify Example 4

Difference of the two alternate interior angles $=64-64=0$. Zero difference is exactly the parallel condition, so $m\parallel n$ ✓.

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Cell E — Perpendicular link

Forecast: Both make $90°$ with $q$. Same relative angle → guess $p\parallel r$, all angles $90°$.

What to notice in the figure: the red horizontal line $q$ acts as the transversal, and the two black vertical lines $p$ and $r$ each cross it. The small square symbols mark the $90°$ right angles at both crossings — identical right angles in the same position are what force $p$ and $r$ to be parallel.

Step 1. At $p\cap q$: angle $=90°$; at $r\cap q$: angle $=90°$. Why this step? That is the definition of $\perp$ — it fixes the angle exactly.

Step 2. These are corresponding angles (same relative position along transversal $q$), and they are equal ($90°=90°$). Why this step? Treat $q$ as the transversal cutting $p$ and $r$; the two right angles occupy matching positions.

Step 3. By the converse of the corresponding-angles axiom, $p\parallel r$. All four angles at each crossing are $90°$. Why this step? Equal corresponding angles force parallelism; and one $90°$ angle forces the other three (each remaining angle is a linear-pair partner or vertically opposite, so also $90°$).

Recall Verify Example 5

Each angle $=90°$; sum around a point $=4\times 90=360°$ ✓; adjacent linear pair $90+90=180°$ ✓; $p\parallel r$ confirmed.

This is exactly the "two lines perpendicular to a third are parallel" property, useful later for triangle constructions (see the triangles note).

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Cell F — Degenerate / limiting case

Forecast: When $t\perp$, every angle should collapse to $90°$. In the flat case there is no genuine crossing.

Step 1 (perpendicular limit). Each of the $8$ angles $=90°$. Why this step? A right-angle transversal makes all created angles right angles; there is nothing left to distinguish "acute" from "obtuse".

Step 2. Alternate interior: $90°=90°$ ✓ (equal, as always). Co-interior: $90°+90°=180°$ ✓ (supplementary, as always). Why this step? The general rules must still hold at the extreme — this is a good sanity check that the theory is consistent.

Step 3 (degenerate $0°$ case). If $t$ coincides with $l_1$, it does not cut $l_1$ at a distinct point on both sides, so it is not a transversal (a transversal must meet the lines at distinct points). Why this step? Checking the boundary tells you the rules simply don't apply — you must reject this input, not force an answer.

Recall Verify Example 6

Perpendicular limit: alternate interior difference $90-90=0$ ✓; co-interior sum $90+90=180$ ✓. Degenerate case: no valid distinct intersection, rules undefined.

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Cell G — Word problem (real world)

Forecast: "West side, between tracks" versus the given "east side" → alternate interior → guess $58°$.

Step 1. Model: tracks $=l_1\parallel l_2$, footbridge $=$ transversal $t$. Given angle $=58°$ interior at the north crossing, east side. Why this step? Turn the physical picture into the standard transversal diagram so the angle rules apply.

Step 2. Wanted angle is interior at the south crossing, west side — opposite side of $t$, between the parallels → alternate interior. Why this step? Identifying position (interior, opposite side) selects the rule.

Step 3. Alternate interior angles are equal, so wanted angle $=58°$. Why this step? The parallel tracks guarantee the Z-shape equality.

Recall Verify Example 7

Answer $58°$. Its co-interior partner (west side, north crossing) would be $180-58=122°$, and $58+122=180°$ ✓ — internally consistent.

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Cell H — Exam twist (bent transversal)

Forecast: A classic trap — the bent line is not one transversal. Draw an auxiliary parallel through $P$. Guess the answer $=35+48=83°$.

What to notice in the figure: the two black parallels $l_1$ and $l_2$ are joined by a bent path that kinks at point $P$. The dashed red line $k$ is the helper line drawn through $P$ parallel to both. It slices the kink into two pieces — the upper piece equals $35°$ (alternate interior with $l_1$) and the lower piece equals $48°$ (alternate interior with $l_2$).

Step 1. Draw line $k$ through $P$ with $k\parallel l_1\parallel l_2$ (the red dashed helper line). Why this step? A single bent line breaks the rules; splitting the kink into two clean transversal angles restores them. The auxiliary parallel is the standard unlocking trick.

Step 2. Upper segment and $l_1$ make $35°$; by alternate interior angles between $l_1$ and $k$, the upper part of the kink (above $k$) $=35°$. Why this step? $l_1\parallel k$ with the upper segment as transversal — Z-shape equality.

Step 3. Lower segment and $l_2$ make $48°$; by alternate interior between $k$ and $l_2$, the lower part of the kink (below $k$) $=48°$. Why this step? Same rule on the other half, since $k\parallel l_2$.

Step 4. Kink angle $=35°+48°=83°$. Why this step? The helper line $k$ splits the bend into two pieces that simply add.

Recall Verify Example 8

Kink $=83°$. Cross-check: the two pieces $35°$ and $48°$ each came from a valid alternate-interior pair; sum $35+48=83°$ ✓.

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Recall gauntlet

For deeper proofs of these converses, see the advanced Euclidean geometry note; for using equal angles to prove triangles match, see the congruence of triangles note.