Pairs of angles

Stand a ray on a straight line and something remarkable happens: the two angles it creates are not free to be anything at all — they are locked together, always adding to $180°$ . Cross two straight lines through a point and four angles appear, but really only two distinct sizes show up, repeated across from each other. This section covers the two facts that govern every such figure in CBSE Class 9 geometry: the linear pair axiom (a ray on a line splits it into two angles summing to $180°$ , and the converse) and the vertically opposite angles theorem (angles across an intersection point are equal, and this can be proved). Together they turn any diagram of rays and crossing lines into a short chain of equations — set one up correctly, and the rest is algebra.

The core insight

A ray standing on a line makes two adjacent angles that sum to $180°$ (linear pair), and two lines crossing make vertically opposite angles that are equal — and the second fact follows from the first by a two-line proof.

The core idea

Linear pair axiom. If a ray $OC$ stands on a line $AB$ , with $O$ on the line and $C$ off it, the straight angle $\angle AOB = 180°$ is split into two adjacent angles, $\angle AOC$ and $\angle BOC$ . These form a linear pair:

\angle AOC + \angle BOC = 180°

where: $\angle AOC$ and $\angle BOC$ are the two angles on either side of ray $OC$ , sharing the vertex $O$ and the arm $OC$ , with their other arms ( $OA$ and $OB$ ) lying along the same straight line. The converse is used just as often: if two adjacent angles at a point add to $180°$ , their non-common arms must lie on a single straight line. Both directions are accepted without proof — this is the axiom itself, the starting assumption everything else builds on.

Vertically opposite angles. When two lines $AB$ and $CD$ cross at $O$ , four angles appear. The two that face each other across $O$ — not sharing an arm — are vertically opposite, and the theorem says they are always equal:

\angle AOC = \angle BOD, \qquad \angle AOD = \angle BOC

where: each pair shares only the vertex $O$ ; the arms of one angle are the reverse rays of the arms of the other. Notice the structure: the linear pair axiom is a starting rule about adjacent angles on a straight line, while the vertically opposite theorem is a derived result about non-adjacent angles at a crossing — derived precisely by applying the linear pair axiom twice.

Figure

Two lines AB and CD crossing at point O with four angles marked and the two vertically opposite pairs shaded in matching colours

Derivation AProving vertically opposite angles are equal

Prove that when two lines $AB$ and $CD$ intersect at a point $O$ , the vertically opposite angles $\angle AOC$ and $\angle BOD$ are equal, using only the linear pair axiom applied to the two rays that stand on the two lines at the point of intersection, with no measurement or assumption of equality made in advance.

Step 1. Ray $OA$ stands on line $CD$ (since $C$ , $O$ , $D$ are collinear). By the linear pair axiom, the two adjacent angles it forms sum to $180°$ :

\angle AOC + \angle AOD = 180° \quad \text{...(i)}

Step 2. Ray $OD$ stands on line $AB$ (since $A$ , $O$ , $B$ are collinear). Again by the linear pair axiom:

\angle AOD + \angle BOD = 180° \quad \text{...(ii)}

Step 3. Both (i) and (ii) equal $180°$ , so their left-hand sides equal each other:

\angle AOC + \angle AOD = \angle AOD + \angle BOD

Step 4. The term $\angle AOD$ appears on both sides and can be cancelled, leaving:

\angle AOC = \angle BOD

This proves one vertically opposite pair equal. The identical argument, now using rays $OC$ and $OB$ standing on lines $AB$ and $CD$ , gives $\angle AOD = \angle BOC$ , proving the second pair.

Worked example

Two lines $AB$ and $CD$ intersect at point $O$ . It is given that $\angle AOC = 3y + 15°$ and its vertically opposite angle $\angle BOD = 5y - 25°$ . Find all four angles formed at $O$ .

Step 1. By the vertically opposite angles theorem, $\angle AOC = \angle BOD$ , so:

3y + 15° = 5y - 25°

Step 2. Solve for $y$ :

15° + 25° = 5y - 3y \implies 40° = 2y \implies y = 20°

Step 3. Substitute back:

\angle AOC = 3(20°) + 15° = 75°, \qquad \angle BOD = 5(20°) - 25° = 75°

Both come out equal, confirming the theorem holds.

Step 4. Ray $OA$ stands on line $CD$ , so $\angle AOC$ and $\angle AOD$ form a linear pair:

\angle AOD = 180° - 75° = 105°

By the vertically opposite angles theorem, $\angle BOC = \angle AOD = 105°$ .

Answer. $\angle AOC = 75°$ , $\angle BOD = 75°$ , $\angle AOD = 105°$ , $\angle BOC = 105°$ .

Competitive stretch

If instead only $\angle AOC = 4x + 10°$ and the adjacent angle $\angle AOD = 2x + 20°$ are given (a linear pair, not vertically opposite), the same two ideas still solve it in a different order: set $(4x+10°)+(2x+20°) = 180°$ , which gives $6x = 150°$ , so $x = 25°$ , $\angle AOC = 110°$ and $\angle AOD = 70°$ — and then $\angle BOD = \angle AOC = 110°$ and $\angle BOC = \angle AOD = 70°$ follow immediately from the vertically opposite theorem, without any further equation.

Common traps

Calling any two angles that sum to $180°$ a "linear pair" — the axiom needs them to be adjacent, formed by a ray standing on a line; two unrelated supplementary angles elsewhere in a figure are not a linear pair.
Confusing adjacent angles (share a vertex and one arm, generally sum to $180°$ when on a line) with vertically opposite angles (share only the vertex, always equal, not $180°$ unless each happens to be $90°$ ).
Assuming all four angles at an intersection are equal — only the two opposite pairs are equal; each adjacent pair among the four is supplementary, not equal.
Skipping the two-linear-pair-and-cancel steps when asked to prove vertically opposite angles are equal — a "prove" question needs the full Step 1 to Step 4 chain, not just the final equality stated.

Ask the tutor: If you have a messy figure with several intersecting lines and can't tell which angle pairs are linear pairs versus vertically opposite, ask the AI tutor to walk through labelling each pair with you.

Why is the linear pair axiom accepted without proof but vertically opposite angles must be proved

The linear pair axiom rests directly on the definition of a straight angle as $180°$ — splitting it into two parts that add back to $180°$ is essentially the definition itself, so the syllabus treats it as a starting assumption (an axiom). The vertically opposite theorem is a step further: it is a new claim about angles that do not obviously look equal just from their definition, so it needs to be derived logically from the axiom, which is exactly what the four-step proof does.

How do I tell linear pair and vertically opposite angles apart quickly in a figure

Check whether the two angles share an arm. If they share both the vertex and one arm, and their other two arms together form a straight line, it is a linear pair (adjacent, sums to $180°$ ). If they share only the vertex, with the arms of one being the exact reverse rays of the other, it is a vertically opposite pair (equal, not adjacent). Sketching the arms lightly and asking is there a shared ray answers this in seconds.

If I know one angle at an intersection of two lines, can I always find the other three

Yes, as long as you know one of the four angles formed, the linear pair axiom and vertically opposite theorem together fix the rest: the angle vertically opposite it is equal, and each of the two angles adjacent to it is its supplement ( $180°$ minus the known angle). This is exactly why these two results are so useful — one measurement plus two rules pins down the whole figure.

Pairs of angles

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads