Lines parallel to the same line

The result. Let $l$, $m$, $n$ be three lines in a plane.

where: $l \parallel m$ means line $l$ is parallel to line $m$ (they never meet, however far extended), and the symbol $\implies$ reads "therefore." Line $m$ is the common line both $l$ and $n$ are compared against.

• Stated without proof. At Class 9 level this result is given as a known fact, not something you are asked to prove formally in an exam — the syllabus explicitly marks it this way. Use it directly wherever it applies.
• Extends to any number of lines. If $l_1 \parallel m$, $l_2 \parallel m$, $l_3 \parallel m$, and so on, then every pair among $l_1, l_2, l_3, \ldots$ is mutually parallel — not just each one to $m$. Several lines can all be tied together through one shared reference line.
• A tool, not a standalone theorem. In practice this result rarely appears as "prove that..."; it shows up mid-problem as a quick justification step: "since both lines are parallel to the given line, they are parallel to each other," letting you move on to the next part of the question.
• Needs a shared line. The result only connects $l$ and $n$ because both are stated parallel to the same line $m$. Two unrelated parallel statements, like $l \parallel m$ and $p \parallel q$, say nothing about $l$ and $p$ unless $m$ and $q$ are linked somehow.

The syllabus does not require a formal proof, but the intuition is worth owning because it makes the result feel obvious rather than memorised. Parallel lines keep a constant perpendicular distance and never meet, no matter how far they are extended. If $l$ never meets $m$, and $m$ never meets $n$, both facts are really statements about direction: $l$ and $m$ point the same way, and so do $m$ and $n$. Since $m$ cannot point two different ways at once, $l$ and $n$ must share that same direction too — and lines pointing the same way, once you rule out them being the same line, never meet. That shared direction is exactly what "parallel" means. The informal proof most students picture makes this concrete with a transversal. Draw a single line $t$ cutting across all three lines $l$, $m$, $n$. Because $l \parallel m$, the transversal makes equal corresponding angles with $l$ and $m$ — this is the standard converse-of-corresponding-angles fact for a transversal on parallel lines. Because $m \parallel n$, the same transversal makes equal corresponding angles with $m$ and $n$. Both of these equal angles are measured at the same position relative to $m$, so the angle the transversal makes with $l$ equals the angle it makes with $n$. Equal corresponding angles with the same transversal means $l \parallel n$. This argument is informal support for intuition, not a Class 9 exam requirement — the result itself is quoted directly.

In a figure, line $AB \parallel$ line $CD$, and line $CD \parallel$ line $EF$. A transversal cuts all three lines. If the angle the transversal makes with $AB$ at its intersection point is $65°$ on a given side, what can be said about lines $AB$ and $EF$, and what is the corresponding angle the transversal makes with $EF$?

Step 1. Apply the "lines parallel to the same line" result. We are given $AB \parallel CD$ and $CD \parallel EF$. Both $AB$ and $EF$ are parallel to the same line $CD$, so by the result, $AB \parallel EF$.

Step 2. Find the angle the transversal makes with $EF$. Since $AB \parallel EF$ from Step 1, and one transversal cuts both, corresponding angles between the transversal and $AB$, and the transversal and $EF$, are equal: the angle with $EF$ equals the angle with $AB$, which is $65°$.

Final answer: $AB \parallel EF$, and the transversal makes a $65°$ corresponding angle with $EF$ as well.

Competitive stretch

Four lines $p$, $q$, $r$, $s$ satisfy $p \parallel q$, $q \parallel r$, and $r \parallel s$. Is it correct that all four lines are mutually parallel?

Step 1. Combine $p \parallel q$ and $q \parallel r$: both $p$ and $r$ are parallel to the same line $q$, so $p \parallel r$.

Step 2. Combine $r \parallel s$ with $p \parallel r$ from Step 1: both $p$ and $s$ are now parallel to the same line $r$, so $p \parallel s$.

Step 3. Check the remaining pair, $q$ and $s$: both are parallel to $r$ (given $q \parallel r$, and $r \parallel s$ reversed gives $s \parallel r$), so $q \parallel s$ as well. Every pair among $p, q, r, s$ is now confirmed parallel. The claim is correct — all four lines are mutually parallel, chained through the shared lines $q$ and $r$.

• Trying to prove this result with a full angle-chase as if a derivation were compulsory — the syllabus marks it "state without proof," so in exams apply it directly as a known fact; the transversal reasoning is for understanding only.
• Assuming the same simple logic carries over to perpendicularity without checking — "lines perpendicular to the same line are parallel to each other" is a true but different statement, and should not be substituted blindly for the parallel-to-parallel case.
• Forgetting the result needs a genuinely shared line — $l \parallel m$ and $p \parallel q$ tells you nothing about $l$ and $p$ unless $m$ and $q$ are connected.
• Checking each pair one by one in a figure with several lines all parallel to one reference line — it is faster to conclude they are all mutually parallel in a single step, as in the competitive-stretch example.

Ask the tutor: if your figure has several lines and multiple "parallel to" statements chained together, ask the AI tutor to help you spot which lines share a common reference line — that is the whole trick to using this result confidently.