Bisect a given angle using compass and straightedge — steps

Question

Given an angle $\angle AOB$ , construct the angle bisector using only a compass and straightedge. Write the steps of construction and justify why it works.

Solution — Step by Step

Place the compass at vertex $O$ (the point of the angle). Draw an arc of any convenient radius that cuts both rays $OA$ and $OB$ . Let this arc meet ray $OA$ at point $P$ and ray $OB$ at point $Q$ .

Why: We need two points equidistant from the vertex, one on each ray, to set up a symmetric construction.

Without changing the compass width, place the compass tip at $P$ and draw an arc in the interior of the angle. Then, keeping the same radius, place the compass tip at $Q$ and draw another arc in the interior. The two arcs should intersect at a point — call it $R$ .

Why: Both arcs have the same radius. Any point on the arc from $P$ is equidistant from $P$ ; any point on the arc from $Q$ is equidistant from $Q$ . The intersection $R$ is equidistant from both $P$ and $Q$ .

Use a straightedge to draw a ray from $O$ through $R$ . This ray $OR$ is the angle bisector of $\angle AOB$ .

The ray $OR$ divides $\angle AOB$ into two equal parts:

\angle AOR = \angle BOR = \frac{1}{2}\angle AOB

You can verify by measuring both angles with a protractor — they should be equal.

Why This Works — Geometric Proof

We need to show that ray $OR$ bisects $\angle AOB$ , i.e., that $\angle AOR = \angle BOR$ .

Consider triangles $\triangle OPR$ and $\triangle OQR$ :

$OP = OQ$ (both are radii of the arc drawn in Step 1)
$PR = QR$ (both are radii of the equal arcs drawn in Step 2)
$OR = OR$ (common side)

By SSS congruence, $\triangle OPR \cong \triangle OQR$ .

Therefore, $\angle POR = \angle QOR$ (corresponding angles in congruent triangles).

Since $P$ is on $OA$ and $Q$ is on $OB$ , this means $\angle AOR = \angle BOR$ . The angle is bisected. $\blacksquare$

In CBSE Class 9, construction questions carry 4 marks: 1 for steps, 2 for accurate construction, 1 for justification. Don’t skip the justification — write the SSS congruence proof in full. Many students draw correctly but lose the justification mark.

Application — Constructing Special Angles

Once you know angle bisection, you can construct a variety of angles:

$90°$ → bisect a straight angle (180°)
$45°$ → bisect a $90°$ angle
$60°$ → construct an equilateral triangle and take one angle
$30°$ → bisect a $60°$ angle
$15°$ → bisect a $30°$ angle

The angle of $75°$ requires combining: $60° + 15°$ or equivalently $90° - 15°$ .

Common Mistake

A very common error is changing the compass radius between the two arcs drawn from $P$ and $Q$ (Step 2). Both arcs must have the same radius — the exact value doesn’t matter, but both must be equal. If you change the radius, the intersection point $R$ will NOT be equidistant from $P$ and $Q$ , and the bisector will be wrong. The easiest way to avoid this: do not adjust the compass between the two arcs in Step 2. Draw from $P$ , then immediately move to $Q$ without touching the compass width.

Question

Solution — Step by Step

Why This Works — Geometric Proof

Application — Constructing Special Angles

Common Mistake

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