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

Top intersection (line $l_1$ meets transversal $t$): $\angle 1$ top-left, $\angle 2$ top-right, $\angle 3$ bottom-right, $\angle 4$ bottom-left. Bottom intersection (line $l_2$ meets $t$): $\angle 5$ top-left, $\angle 6$ top-right, $\angle 7$ bottom-right, $\angle 8$ bottom-left. Interior angles = the four "in the corridor" ones: $\angle 3,\angle 4$ (top) and $\angle 5,\angle 6$ (bottom). Rules (only when $l_1\parallel l_2$): corresponding equal ($\angle 1=\angle 5$ etc.), alternate interior equal ($\angle 3=\angle 5$, $\angle 4=\angle 6$), co-interior supplementary ($\angle 4+\angle 5=180°$, $\angle 3+\angle 6=180°$).

What we look at: $\angle 2$ is top-right at the top intersection; $\angle 6$ is top-right at the bottom intersection. Same relative position, same side of the transversal. Why: Same relative position at the two intersections = the corresponding pattern (the "F-shape"). Corresponding angles are equal when the lines are parallel. Answer: Corresponding angles, and they are equal: $\angle 2=\angle 6$.

$\angle 4$ is bottom-left (top intersection) and $\angle 5$ is top-left (bottom intersection). Both are interior (in the corridor), and they sit on the same (left) side of the transversal. Same side + interior = co-interior ("C-shape"). Co-interior angles are supplementary. Answer: $\angle 4+\angle 5=180°$.

Vertically opposite to $\angle 5$ (top-left, bottom intersection) is the diagonally-across angle at that same intersection: $\angle 7$ (bottom-right). Corresponding to $\angle 4$ (bottom-left, top) is the bottom-left angle below it: $\angle 8$. Hmm — those are different, so re-read: we want ONE angle satisfying both. Vertically opposite $\angle 5$ is $\angle 7$. Corresponding to $\angle 4$ is $\angle 8$. They are not equal, so no single angle is both — but if we instead ask "corresponding to $\angle 1$" that is $\angle 5$, whose vertical opposite is $\angle 7$. Answer: The angle vertically opposite $\angle 5$ is ==$\angle 7$==; the angle corresponding to $\angle 4$ is $\angle 8$. There is no single angle that is both. (This one is a trap — see below.)

$\angle 2$: $\angle 1$ and $\angle 2$ sit side-by-side on line $l_1$, forming a straight line (a linear pair). So $\angle 2=180°-118°=62°$. $\angle 5$: corresponding to $\angle 1$, so equal: $\angle 5=118°$. $\angle 6$: corresponding to $\angle 2$, so $\angle 6=62°$. (Check: $\angle 5+\angle 6=180°$, a linear pair on $l_2$. ✓) Answers: $\angle 2=62°$, $\angle 5=118°$, $\angle 6=62°$.

Why this equation: co-interior angles are supplementary, so their measures must add to $180°$. $(2x+10)+(3x-30)=180$ $5x-20=180⟹5x=200⟹x=40.$ Then $\angle 4=2(40)+10=90°$ and $\angle 5=3(40)-30=90°$. Answers: $x=40$, $\angle 4=90°$, $\angle 5=90°$. (Both right angles — the transversal happened to be perpendicular here.)

$\angle 3$ and $\angle 6$ are on opposite sides, both interior — but they are NOT the alternate-interior pair (those are $\angle 3=\angle 5$ and $\angle 4=\angle 6$). $\angle 3$ and $\angle 6$ are the co-interior pair $\angle 3+\angle 6=180°$. So $\angle 6=180°-47°=133°$. Answer: $\angle 6=133°$.

Link them: $\angle 4$ (bottom-left, top) corresponds to $\angle 8$ (bottom-left, bottom): $\angle 4=\angle 8$. And $\angle 7,\angle 8$ form a linear pair on $l_2$: $\angle 7+\angle 8=180°$. So $\angle 4+\angle 7=180°$. $3y+(y+80)=180⟹4y=100⟹y=25.$ Then $\angle 4=3(25)=75°$. Check: $\angle 7=25+80=105°$, and $75+105=180°$. ✓ Answers: $y=25$, $\angle 4=75°$.

Why one equation: $\angle 3$ and $\angle 5$ are the alternate-interior pair, so they are equal. $4a-5=2a+45⟹2a=50⟹a=25.$ Common angle $=2(25)+45=95°$. Check with the other form: $4(25)-5=95°$. ✓ Answers: $a=25$, angle $=95°$.

Trick: draw a helper line through $P$ parallel to $A$ and $B$ (dashed in the figure). It splits the bend angle into two. Top part: the $35°$ at $A$ and the upper piece at $P$ are alternate interior angles across the helper line $\Rightarrow$ upper piece $=35°$. Bottom part: the $42°$ at $B$ and the lower piece at $P$ are alternate interior angles $\Rightarrow$ lower piece $=42°$. Bend angle $=35°+42°=77°$. Answer: $\angle P=77°$.

Why $p\parallel r$: both are perpendicular to $q$, so they are corresponding-equal ($90°=90°$) across $q$, hence $p\parallel r$ (parent note, perpendicular property 2). Then: since $p\parallel r$, the transversal $t$ makes equal corresponding angles with each. So $t$ meets $r$ at the same $58°$. Answer: $58°$.

Let $\angle 4=4k$, $\angle 5=5k$. Co-interior $\Rightarrow 4k+5k=180\Rightarrow 9k=180\Rightarrow k=20$. So $\angle 4=80°$, $\angle 5=100°$. For the transversal to be perpendicular, every one of the 8 angles would be $90°$; here they are not, so not perpendicular. Answers: $\angle 4=80°$, $\angle 5=100°$, not perpendicular.

Inside triangle $EFG$ the three interior angles sum to $180°$ (angle-sum property). $\angle FEG+\angle EFG+\angle EGF=180°$ $50°+60°+\angle EGF=180°⟹\angle EGF=70°.$ Answer: $\angle EGF=70°$.

Setup: transversal $t$ meets lines $m$ and $n$ at $E$ and $F$. Given $\angle 3=\angle 5$ (alternate interior).

1. Suppose, for contradiction, $m\nparallel n$. Then $m$ and $n$ meet at some point $P$, forming triangle $EFP$.
2. In that triangle, $\angle 3$ is an exterior angle at $E$ (or $\angle 5$ at $F$), and by the exterior-angle theorem it is strictly greater than the remote interior angle at the other vertex.
3. But that remote interior angle equals the given angle (they are the same alternate-interior measure), so we would need $\angle 3>\angle 3$ — impossible.
4. Contradiction. Hence $m$ and $n$ do not meet: $m\parallel n$. $■$ (Compare with congruence proofs — same "assume the opposite, derive a contradiction" engine.)

Take co-interior pair $\angle 4$ (top) and $\angle 5$ (bottom).

1. $\angle 4$ corresponds to $\angle 8$ (both bottom-left, at the two crossings) $\Rightarrow \angle 4=\angle 8$ (corresponding axiom, $l_1\parallel l_2$).
2. $\angle 5$ and $\angle 8$ are a linear pair on line $l_2$ (top-left + bottom-left share vertex $F$ and lie along $l_2$) $\Rightarrow \angle 5+\angle 8=180°$.
3. Substitute $\angle 8=\angle 4$: $\angle 5+\angle 4=180°$. $■$

What happens: if the third line is parallel to $l_1$ and $l_2$, it never meets them at distinct points — it either lies on top of one of them or runs alongside forever. There are no intersection points, so no angles are formed. Is it a transversal? No. By definition a transversal must cut two or more lines at distinct points. A line parallel to both fails this, so the whole angle machinery does not apply. Answer: degenerate — no crossings, no angles, not a transversal. (This is the boundary that the definition deliberately excludes.)

Let the transversal make an acute angle $\theta$ with $l_1$ where $\theta$ grows from near $0°$ up to $90°$.

• Then $\angle 4=\theta$ and its co-interior partner $\angle 5=180°-\theta$ (they must sum to $180°$ at every instant).
• As $\theta \to 0°$ (nearly parallel): $\angle 4\to 0°$, $\angle 5\to 180°$ — the corridor angles flatten out.
• At $\theta =90°$ (perpendicular): $\angle 4=\angle 5=90°$ — all eight angles equal, the symmetric case.
• As $\theta \to 90°$ from below, $\angle 4$ and $\angle 5$ both approach $90°$ from opposite sides. At every value of $\theta$, $\angle 4+\angle 5=\theta +(180°-\theta )=180°$. The supplementary relationship is preserved for the whole rotation — it never depends on the particular tilt, only on parallelism. Answer: $\angle 4=\theta$, $\angle 5=180°-\theta$; always supplementary; equal ($90°$) exactly when the transversal is perpendicular.

Which pair is supplementary, not equal, for parallel lines? ::: Co-interior (same-side interior) angles. A line parallel to both "target" lines — is it a transversal? ::: No; it makes no distinct intersection points. Two lines both perpendicular to a third are... ::: parallel to each other.