Geometric Constructions — Bisectors, Triangles, Tangents

Why Constructions Matter

Geometric constructions are about building precise geometric figures using only a compass and a straightedge — no measurements, no protractor. This is one of those chapters where understanding the why behind each step separates students who score full marks from those who lose marks on “explain the construction” type questions.

For CBSE Classes 9 and 10, constructions carry 3-5 marks in the board exam. The questions are predictable — bisectors, triangle constructions, and tangents from external points. Master these three categories and you’re set.

graph TD
    A[Construction Problem] --> B{What type?}
    B -->|Bisect angle/line| C[Bisector Construction]
    B -->|Build triangle| D[Triangle Construction]
    B -->|Draw tangent to circle| E[Tangent Construction]
    C --> C1[Equal arcs from vertex]
    C1 --> C2[Arcs intersect → join to vertex]
    D --> D1{Given data?}
    D1 -->|Base + angles| D2[ASA Construction]
    D1 -->|Sides only| D3[SSS Construction]
    D1 -->|Two sides + angle| D4[SAS Construction]
    E --> E1{Point position?}
    E1 -->|On circle| E2[Radius → perpendicular]
    E1 -->|Outside circle| E3[Midpoint method]

Key Terms & Definitions

Compass — The tool that draws arcs and circles. It maintains a fixed radius once set.

Straightedge — A ruler without markings. We use it only to draw straight lines through given/constructed points, not to measure lengths.

Angle Bisector — A ray that divides an angle into two equal parts. Every point on the bisector is equidistant from the two arms of the angle.

Perpendicular Bisector — A line that is perpendicular to a segment and passes through its midpoint. Every point on this line is equidistant from the two endpoints.

Tangent — A line that touches a circle at exactly one point. At the point of tangency, the tangent is perpendicular to the radius.

Methods — Step by Step

Construction 1: Bisecting a Line Segment

Set the compass to any radius greater than half the segment length. With centre at one endpoint, draw an arc above and below the segment. Repeat from the other endpoint with the same radius.

The two arcs intersect at two points (one above, one below the segment). Draw a straight line through these two points. This line is the perpendicular bisector — it passes through the midpoint and is perpendicular to the segment.

Why does this work? Each arc is a set of points equidistant from one endpoint. The intersection points are equidistant from both endpoints — and the locus of such points is exactly the perpendicular bisector.

Construction 2: Bisecting an Angle

With the vertex as centre, draw an arc that intersects both arms of the angle. Call these intersection points $P$ and $Q$ .

With $P$ and $Q$ as centres and equal radius, draw two arcs that intersect each other at a point $R$ inside the angle.

Draw the ray from the vertex through $R$ . This is the angle bisector.

Construction 3: Triangle Given Base, Base Angle, and Sum of Other Two Sides

This is a classic CBSE Class 9 construction.

Draw segment $BC$ of the given length.

At $B$ , construct the given angle. Draw ray $BX$ making this angle with $BC$ .

Cut off $BD$ on ray $BX$ where $BD = AB + AC$ (the given sum of the other two sides).

Join $DC$ . Construct the perpendicular bisector of $DC$ . Let it intersect $BD$ at $A$ . Triangle $ABC$ is the required triangle.

Why the perpendicular bisector of $DC$ ? Any point on the perpendicular bisector of $DC$ is equidistant from $D$ and $C$ . So $A$ (on this bisector) satisfies $AD = AC$ . Since $BD = AB + AD = AB + AC$ , our condition is met.

Construction 4: Tangent to a Circle from an External Point

This is the most important Class 10 construction.

Let $O$ be the centre and $P$ the external point. Draw segment $OP$ .

Construct the perpendicular bisector of $OP$ to find its midpoint $M$ .

With $M$ as centre and $MO$ as radius, draw a circle (or semicircle). This circle intersects the original circle at points $T_1$ and $T_2$ .

Join $PT_1$ and $PT_2$ . These are the two tangent lines from $P$ to the circle.

Why does this work? $T_1$ lies on the semicircle, so angle $OT_1P = 90°$ (angle in semicircle). Since $OT_1$ is a radius and $OT_1 \perp PT_1$ , the line $PT_1$ must be a tangent. This reasoning is often asked as a “justify the construction” question — worth 1-2 marks.

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE Class 9)

Construct a triangle $ABC$ where $BC = 7$ cm, $\angle B = 75°$ , and $AB + AC = 13$ cm.

Follow Construction 3 above. Draw $BC = 7$ cm. At $B$ , construct $75°$ (construct $60° + 15°$ , or use $60°$ and bisect to get $90°$ , then bisect the $30°$ gap). Mark $BD = 13$ cm on the ray. Join $DC$ , bisect it, get $A$ . Verify with compass that $AB + AC = 13$ cm.

Example 2 (Medium — CBSE Class 10)

Draw a circle of radius 4 cm. From a point 7 cm from the centre, construct two tangents to the circle. Measure the tangent length.

Follow Construction 4. After constructing, measure $PT_1$ . By Pythagoras: $PT_1 = \sqrt{7^2 - 4^2} = \sqrt{33} \approx 5.74$ cm. Your measured value should be close to this.

Example 3 (Medium — CBSE Class 10)

Construct a triangle similar to a given triangle $ABC$ with sides in ratio $3:5$ .

Draw triangle $ABC$ first. From vertex $A$ , draw a ray $AX$ at an acute angle to $AB$ . Mark 5 equal divisions on $AX$ . Join the 5th division to $B$ . Through the 3rd division, draw a line parallel to this join — it meets $AB$ at $B'$ . From $B'$ , draw a line parallel to $BC$ meeting $AC$ at $C'$ . Triangle $AB'C'$ is similar to $ABC$ with ratio $3:5$ .

Exam-Specific Tips

CBSE Class 9: Constructions carry 3-4 marks. The most frequently asked types are: (1) triangle with base + base angle + sum/difference of other sides, and (2) triangle with base + base angle + perimeter. Always write the steps of construction alongside the figure.

CBSE Class 10: The big ones are tangent constructions and similar triangle constructions. In 2023 and 2024, tangent from external point appeared in almost every set. The “justify the construction” part is often asked for 1-2 extra marks — learn the angle-in-semicircle reasoning.

Common Mistakes to Avoid

Mistake 1 — Not keeping the compass width constant during a step. When drawing arcs from $P$ and $Q$ to find the bisector, both arcs must use the same radius. If the compass slips, the intersection point shifts and the bisector is wrong.

Mistake 2 — Forgetting construction arcs in the final figure. CBSE examiners look for the arc marks. If you erase them or don’t show them, you lose marks even if the triangle is correct. Keep all construction lines visible.

Mistake 3 — Confusing sum vs. difference constructions. When the sum of two sides is given, mark the sum on the ray. When the difference is given, the construction is slightly different — you mark the difference and the perpendicular bisector goes on the other segment. Mixing these up gives a completely wrong triangle.

Mistake 4 — Drawing tangents by eye instead of using the semicircle method. A tangent “looks” perpendicular to the radius, but drawing it freehand loses marks. Always use the midpoint-semicircle construction to get exact tangent points.

Mistake 5 — Not labelling all points. In board exams, every point must be labelled as mentioned in the question. Missing labels cost marks.

Practice Questions

Q1. Construct an angle of $60°$ using compass and straightedge. Then bisect it to get $30°$ .

Draw a ray. With the endpoint as centre, draw an arc. From the intersection of the arc with the ray, draw another arc with the same radius. The intersection gives $60°$ . Now bisect: from the two points on the arc, draw equal arcs to find the bisector. This gives $30°$ .

Q2. Construct a triangle where $BC = 8$ cm, $\angle B = 45°$ , and $AB - AC = 3.5$ cm.

Draw $BC = 8$ cm. At $B$ , construct $45°$ . Since $AB > AC$ , mark $BD = 3.5$ cm on ray $BX$ . Join $DC$ . Construct perpendicular bisector of $DC$ meeting $BX$ at $A$ . Join $AC$ . Triangle $ABC$ is the answer.

Q3. Draw a circle of radius 3 cm. Construct a tangent at a point on the circle.

Draw circle with centre $O$ and radius 3 cm. Mark a point $T$ on the circle. Draw radius $OT$ . At $T$ , construct a perpendicular to $OT$ — this is the tangent. Use the standard perpendicular construction (arcs from $T$ on the radius line, then arcs from those points).

Q4. Construct a circle of radius 5 cm. From a point 10 cm from the centre, draw two tangents. Find the tangent length.

Follow the external tangent construction. Tangent length $= \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ cm.

Q5. Construct a triangle $PQR$ where $QR = 6$ cm, $\angle Q = 60°$ , and the perimeter is 16 cm.

Perimeter = 16, so $PQ + PR = 16 - 6 = 10$ cm. This reduces to the “sum of two sides” construction. Draw $QR = 6$ cm, angle $60°$ at $Q$ , mark 10 cm on the ray, join to $R$ , bisect to find $P$ .

Q6. Draw a line segment of 7.8 cm and divide it in ratio $5:8$ .

Draw $AB = 7.8$ cm. Draw ray $AX$ at acute angle. Mark 13 (= 5 + 8) equal divisions. Join the 13th to $B$ . Through the 5th division, draw a line parallel to this join — it meets $AB$ at the required point.

Q7. Construct a triangle similar to triangle $ABC$ with scale factor $\frac{2}{3}$ .

Draw triangle $ABC$ . From $A$ , draw ray at acute angle to $AB$ . Mark 3 equal parts. Join 3rd to $B$ . Through 2nd, draw parallel to this line meeting $AB$ at $B'$ . From $B'$ , draw parallel to $BC$ meeting $AC$ at $C'$ . Triangle $AB'C'$ is the answer.

Q8. Construct an equilateral triangle of side 5 cm and then construct a triangle similar to it with sides $\frac{2}{3}$ of the original.

First construct equilateral triangle: draw base 5 cm, arcs of 5 cm from both ends, intersection gives the third vertex. Then apply the similar triangle construction with ratio 2:3 as in Q7.

FAQs

Why can’t we use a protractor in constructions?

Classical geometric constructions use only compass and straightedge. This tradition comes from ancient Greek mathematics — the idea is to build figures from first principles (equal distances and straight lines) without relying on measuring tools. In board exams, using a protractor in a construction question can cost marks.

How many tangents can be drawn from a point to a circle?

If the point is outside the circle: 2 tangents. If on the circle: 1 tangent. If inside the circle: 0 tangents.

What is the difference between construction and drawing?

A “drawing” can use any tool (ruler, protractor, set squares). A “construction” uses only compass and straightedge, and every step must be justifiable geometrically. Board exams specifically ask for constructions — so follow the rules.

How do we construct a 90-degree angle?

Construct a $60°$ angle first. Then construct a $120°$ angle (two $60°$ arcs). Bisect the angle between $60°$ and $120°$ to get $90°$ . Alternatively, construct the perpendicular bisector of any segment — the angles at the midpoint are $90°$ .

Why does the angle in a semicircle equal 90 degrees?

This is Thales’ theorem. If $AB$ is a diameter and $C$ is any point on the circle (not on $AB$ ), then angle $ACB = 90°$ . This is the key reason the tangent construction from an external point works — we create a right angle at the point of tangency.

Can we construct any angle with compass and straightedge?

No. We can construct angles that are multiples of $15°$ (since we can construct $60°$ and bisect). Some angles like $1°$ or $20°$ cannot be constructed with compass and straightedge alone — this was proven by mathematicians using algebra.