Read each topic in order. Study the theorem cards and the standard figures first, then work through the solved examples on paper with the solution covered. Geometry is best learned by drawing: make a free-hand sketch before writing the proof. Finally, attempt the exam-style questions before looking at the answers. Questions marked ICSE Board are verified questions from CISCE board papers and are reused here exactly for practice.
Topic 1 – Similarity
A. Learning Objectives
After studying this topic you should be able to:
- State the meaning of similar figures and the conditions for two triangles to be similar.
- Use the AA, SSS and SAS similarity criteria to prove two triangles similar.
- Write correct correspondence and use the fact that corresponding sides of similar triangles are proportional.
- Apply the result that the ratio of the areas of two similar triangles equals the square of the ratio of corresponding sides.
- State and apply the Basic Proportionality Theorem (Thales’ theorem) and its converse.
- Use the angle-bisector theorem for a triangle.
B. Basic Concepts and Definitions
Two figures are similar if they have the same shape but not necessarily the same size. One is an exact scaled copy of the other. In similar polygons, corresponding angles are equal and corresponding sides are in the same ratio.
Two triangles are similar if any one of the following holds:
- AA (or AAA): two pairs of corresponding angles are equal.
- SSS: the three pairs of corresponding sides are in the same ratio.
- SAS: one pair of corresponding angles is equal and the two pairs of sides including them are in the same ratio.
We write \(\triangle ABC \sim \triangle PQR\), where the order of letters shows the correspondence: \(\angle A = \angle P\), \(\angle B = \angle Q\), \(\angle C = \angle R\).
If a line is drawn parallel to one side of a triangle and cuts the other two sides, then it divides those two sides in the same ratio. In \(\triangle ABC\), if \(DE \parallel BC\) with \(D\) on \(AB\) and \(E\) on \(AC\), then \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\). The converse is also true.
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. In \(\triangle ABC\), if \(AD\) bisects \(\angle A\) and meets \(BC\) at \(D\), then \(\dfrac{BD}{DC} = \dfrac{AB}{AC}\).
C. Important Formulas and Theorems
D. Step-by-Step Solved Examples
Level 1 Using the Basic Proportionality Theorem
In \(\triangle ABC\), \(DE \parallel BC\), with \(AD = 3\text{ cm}\), \(DB = 5\text{ cm}\) and \(AE = 4\text{ cm}\). Find \(EC\).
Level 2 Ratio of areas of similar triangles
Two similar triangles have corresponding sides in the ratio \(2 : 3\). If the area of the smaller triangle is \(24\text{ cm}^2\), find the area of the larger triangle.
Level 3 Basic Proportionality with an unknown
In \(\triangle ABC\), \(DE \parallel BC\) with \(AD = x\), \(DB = x - 2\), \(AE = x + 2\) and \(EC = x - 1\). Find \(x\).
E. Exam-Style Questions
In \(\triangle ABC\) and \(\triangle PQR\), \(AD\) and \(PS\) are the bisectors of \(\angle A\) and \(\angle P\) respectively. If \(\triangle ABC \sim \triangle PQR\), then \(\triangle ABD\) is similar to which triangle?
Show answer
Since the triangles are similar, \(\angle A = \angle P\), so \(\angle BAD = \angle QPS\). Also \(\angle B = \angle Q\). By AA, \(\triangle ABD \sim \triangle PQS\).
Two similar triangles have areas \(49\text{ cm}^2\) and \(81\text{ cm}^2\). Find the ratio of their corresponding sides.
Show answer
Ratio of sides \(= \sqrt{\dfrac{49}{81}} = \dfrac{7}{9}\). So the corresponding sides are in the ratio \(7 : 9\).
In \(\triangle ABC\), \(AB = 8\text{ cm}\), \(AC = 6\text{ cm}\) and \(AD\) bisects \(\angle A\), meeting \(BC\) at \(D\). If \(BD = 4\text{ cm}\), find \(DC\).
Show answer
By the angle-bisector theorem, \(\dfrac{BD}{DC} = \dfrac{AB}{AC}\): \(\dfrac{4}{DC} = \dfrac{8}{6} = \dfrac{4}{3}\), so \(DC = 3\text{ cm}\).
In \(\triangle ABC\), a line \(DE\) parallel to \(BC\) cuts \(AB\) at \(D\) and \(AC\) at \(E\). If \(AD = 4\text{ cm}\), \(AB = 10\text{ cm}\) and \(AE = 6\text{ cm}\), find \(AC\).
Show answer
\(\dfrac{AD}{AB} = \dfrac{AE}{AC}\): \(\dfrac{4}{10} = \dfrac{6}{AC}\), so \(4 \times AC = 60\) and \(AC = 15\text{ cm}\).
In \(\triangle ABC\), \(D\) lies on \(AB\) and \(E\) on \(AC\) such that \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\). Prove that \(DE \parallel BC\).
Show answer
Given \(\frac{AD}{DB} = \frac{AE}{EC}\). By the converse of the Basic Proportionality Theorem, the line joining \(D\) and \(E\) that divides the two sides in the same ratio is parallel to the third side, so \(DE \parallel BC\).
\(\triangle ABC \sim \triangle DEF\). If \(AB = 6\text{ cm}\), \(DE = 9\text{ cm}\) and the area of \(\triangle ABC = 20\text{ cm}^2\), find the area of \(\triangle DEF\).
Show answer
\(\dfrac{\text{Area}(ABC)}{\text{Area}(DEF)} = \left(\dfrac{6}{9}\right)^2 = \dfrac{4}{9}\). So \(\dfrac{20}{\text{Area}(DEF)} = \dfrac{4}{9}\), giving \(\text{Area}(DEF) = \dfrac{20 \times 9}{4} = 45\text{ cm}^2\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Writing triangles in the wrong order, e.g. \(\triangle ABC \sim \triangle QRP\). | Always match corresponding vertices: if \(\angle A = \angle P\), write \(\triangle ABC \sim \triangle PQR\) so the letters line up. |
| Equating non-corresponding sides in a proportion. | Proportional sides must be the pair that lie between the equal angles. Use the order of the letters in the similarity statement. |
| Using the ratio of sides directly for the ratio of areas. | The area ratio is the square of the side ratio: \(\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2\). |
| Applying the Basic Proportionality Theorem without first stating \(DE \parallel BC\). | The theorem needs the line to be parallel to a side; quote the parallel condition (or its converse) explicitly. |
| Mixing up the angle-bisector ratio: using \(\frac{BD}{DC} = \frac{AC}{AB}\). | The correct result is \(\frac{BD}{DC} = \frac{AB}{AC}\) (side opposite the angle is split in the ratio of the sides forming the angle). |
G. Exam Tips
- Draw a clear, labelled figure first – nearly every similarity proof earns method marks for the correct diagram and correspondence.
- State the criterion you are using (AA, SSS or SAS) in words before writing the conclusion.
- In ratio-of-area questions, write the square of the side ratio before substituting numbers.
- For BPT problems with unknowns, form the proportion first and cross-multiply carefully.
H. Quick Revision Sheet – Similarity
- Similar triangles: equal angles, proportional sides; criteria AA, SSS, SAS.
- \(\dfrac{\text{Area}_1}{\text{Area}_2} = \left(\dfrac{\text{corresponding sides}}{}\right)^2 = k^2\).
- BPT: \(DE \parallel BC \Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC}\); the converse is also true.
- Angle-bisector theorem: \(\dfrac{BD}{DC} = \dfrac{AB}{AC}\).
- Ratio of perimeters of similar triangles \(=\) ratio of corresponding sides.
Topic 2 – Loci
A. Learning Objectives
After studying this topic you should be able to:
- Explain what a locus is (the path traced by a point moving under a given condition).
- Recall the standard loci for a point, a line, two points, two intersecting lines and two parallel lines.
- Construct loci accurately using a ruler and compasses only.
- Find the intersection of two loci to locate a point satisfying two conditions at once.
B. Basic Concepts and Definitions
A locus is the set of all points that satisfy a given geometrical condition. It is the complete path traced by a point moving so that the condition always holds. Every point on the locus satisfies the condition, and every point satisfying the condition lies on the locus.
- The locus of a point at a fixed distance \(r\) from a fixed point \(O\) is a circle with centre \(O\) and radius \(r\).
- The locus of a point at a fixed distance \(d\) from a fixed line is a pair of parallel lines, one on each side of the line, at distance \(d\).
- The locus of a point equidistant from two fixed points \(A\) and \(B\) is the perpendicular bisector of \(AB\).
- The locus of a point equidistant from two intersecting lines is the pair of angle bisectors of the angles formed.
- The locus of a point equidistant from two parallel lines is a single line parallel to and midway between them.
- The locus of a point equidistant from the two arms of an angle is the bisector of that angle.
Note: the locus of a point at a fixed distance from a line segment (as opposed to a full line) is the pair of parallel segments together with two semicircles at the ends.
C. Important Constructions (Loci)
D. Step-by-Step Solved Examples
Level 1 Locus at a fixed distance
Describe the locus of a point \(P\) that is always \(3\text{ cm}\) from a fixed point \(O\).
Level 2 Combining two loci
Find the point \(P\) that is equidistant from two fixed points \(A\) and \(B\), and also \(4\text{ cm}\) from \(A\).
Level 3 Locus equidistant from two lines
Two straight lines intersect at \(O\), forming an angle. Describe the locus of a point inside the angle that is equidistant from both lines.
E. Exam-Style Questions
The locus of a point at a fixed distance from a given straight line is:
Show answer
A pair of straight lines parallel to the given line, one on each side, at that fixed distance.
Describe the locus of a point equidistant from the two ends of a line segment \(AB\).
Show answer
The perpendicular bisector of \(AB\).
A point \(P\) moves so that it is always \(2\text{ cm}\) from a fixed line \(l\). Sketch the locus.
Show answer
Two lines parallel to \(l\), one on each side, each \(2\text{ cm}\) from \(l\).
Construct the locus of a point that is equidistant from \(A(0,0)\) and \(B(6,0)\), and also \(5\text{ cm}\) from \(A\). How many such points are there?
Show answer
The perpendicular bisector of \(AB\) is the vertical line \(x = 3\). The circle centre \(A\) radius \(5\) meets it where \(3^2 + y^2 = 25\), i.e. \(y = \pm 4\). So there are two points: \((3,4)\) and \((3,-4)\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Describing the locus at a fixed distance from a line as a single line. | It is two lines, one on each side of the given line. |
| Confusing “equidistant from two points” with “equidistant from two lines”. | Two points → perpendicular bisector of the joining segment; two lines → angle bisector(s). |
| Forgetting the two angle bisectors for two intersecting lines. | Two intersecting lines form four angles; their bisectors form two straight lines (internal and external). |
| Giving a sketch instead of a construction. | Use ruler and compasses and leave the construction arcs; loci questions are marked for accurate construction. |
G. Exam Tips
- Learn the six standard loci by heart – they cover almost every exam question.
- When a question has two conditions, draw both loci and mark their intersections.
- State whether a fixed distance is from a point, a line or a segment, as the locus changes accordingly.
H. Quick Revision Sheet – Loci
- Fixed distance from a point → circle.
- Fixed distance from a line → pair of parallel lines.
- Equidistant from two points → perpendicular bisector.
- Equidistant from two intersecting lines → the angle bisectors.
- Equidistant from two parallel lines → the mid-line.
Topic 3 – Circles
A. Learning Objectives
After studying this topic you should be able to:
- Recall the parts of a circle: centre, radius, diameter, chord, arc, sector, segment, semicircle.
- Use the theorem: the angle subtended by an arc at the centre is double the angle at any point on the remaining part.
- Use the results: angles in the same segment are equal, and the angle in a semicircle is a right angle.
- Apply the cyclic quadrilateral property (opposite angles sum to \(180^\circ\)) and its exterior-angle form.
- Use the perpendicular-from-the-centre result and the relation between equal chords and equal arcs.
B. Basic Concepts and Definitions
- Chord: a segment joining two points on the circle.
- Arc: a part of the circle; a minor arc is shorter, a major arc longer.
- Diameter: the longest chord, passing through the centre.
- Sector: region between two radii and an arc. Segment: region between a chord and an arc.
- The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circle.
- Angles in the same segment are equal.
- The angle in a semicircle is a right angle; conversely, if a chord subtends \(90^\circ\) at a point on the circle, it is a diameter.
A quadrilateral whose four vertices lie on a circle. Its opposite angles are supplementary (\(\angle A + \angle C = 180^\circ\), \(\angle B + \angle D = 180^\circ\)). Also, an exterior angle equals the interior opposite angle.
The perpendicular from the centre to a chord bisects the chord. Equal chords are equidistant from the centre, and chords equidistant from the centre are equal. Equal chords subtend equal angles at the centre (and cut off equal arcs).
C. Important Theorems
D. Step-by-Step Solved Examples
Level 1 Centre and circumference angles
In a circle with centre \(O\), the angle subtended by a chord \(AB\) at the centre is \(120^\circ\). Find the angle subtended by \(AB\) at a point on the major arc.
Level 2 Cyclic quadrilateral
\(ABCD\) is a cyclic quadrilateral. If \(\angle A = 75^\circ\), find \(\angle C\).
Level 3 Combining two circle theorems
\(AB\) is a diameter of a circle. \(C\) is a point on the circle such that \(\angle ABC = 35^\circ\). Find \(\angle BAC\).
E. Exam-Style Questions
In a circle with centre \(O\), \(\angle AOB = 80^\circ\). Find the angle subtended by \(AB\) at a point on the major arc.
Show answer
Angle at circumference \(= \dfrac{80^\circ}{2} = 40^\circ\).
\(P\), \(Q\), \(R\) are points on a circle such that \(\angle PRQ = 50^\circ\). Find \(\angle POQ\), where \(O\) is the centre and \(P\), \(Q\) subtend the same arc.
Show answer
Angle at centre \(= 2 \times\) angle at circumference \(= 100^\circ\).
\(ABCD\) is a cyclic quadrilateral with \(\angle B = 3x\) and \(\angle D = 2x\). Find \(x\) and the two angles.
Show answer
\(\angle B + \angle D = 180^\circ\), so \(5x = 180^\circ\) and \(x = 36^\circ\). Hence \(\angle B = 108^\circ\) and \(\angle D = 72^\circ\).
In a circle, \(AB\) is a chord of length \(16\text{ cm}\) and the radius is \(10\text{ cm}\). Find the distance of \(AB\) from the centre.
Show answer
The perpendicular from the centre \(O\) bisects \(AB\), so \(AM = 8\text{ cm}\). In right \(\triangle OMA\): \(OM^2 = OA^2 - AM^2 = 100 - 64 = 36\), so \(OM = 6\text{ cm}\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Halving the angle at the centre instead of doubling the angle at the circumference. | Centre angle \(= 2 \times\) circumference angle. Rearrange carefully depending on which is given. |
| Using the minor-arc angle when the point is on the minor arc. | The centre angle \(2\angle APB\) refers to the arc not containing \(P\). Check which arc the point lies on. |
| Adding the wrong pair of angles in a cyclic quadrilateral. | It is the opposite angles that are supplementary, not adjacent ones. |
| Assuming a chord is a diameter without justification. | A chord is a diameter only if the angle subtended at the circumference is \(90^\circ\) (or it passes through the centre). |
| Forgetting that the perpendicular from the centre bisects the chord. | Use it to set up equal half-chords before applying Pythagoras. |
G. Exam Tips
- Mark the centre and join it to relevant points; radii help form isosceles triangles.
- Write the theorem you are using by name before the calculation.
- For chord-distance problems, drop the perpendicular from the centre and use Pythagoras with the half-chord.
- Check which arc or segment a given point lies on before choosing the angle relationship.
H. Quick Revision Sheet – Circles
- \(\angle\) at centre \(= 2 \times \angle\) at circumference (same arc).
- Angles in the same segment are equal; angle in a semicircle is \(90^\circ\).
- Cyclic quadrilateral: opposite angles sum to \(180^\circ\).
- Perpendicular from the centre bisects the chord.
- Equal chords are equidistant from the centre and subtend equal angles at the centre.
Topic 4 – Tangents and Intersecting Chords
A. Learning Objectives
After studying this topic you should be able to:
- Define a tangent and use the result that the tangent is perpendicular to the radius at the point of contact.
- Use the fact that tangents from an external point are equal in length.
- Apply the alternate segment theorem (tangent–chord angle).
- Use the intersecting chords result \(PA \cdot PB = PC \cdot PD\).
- Use the tangent–secant (power of a point) result \(PT^2 = PA \cdot PB\).
B. Basic Concepts and Definitions
A straight line that touches the circle at exactly one point, called the point of contact. At the point of contact, the tangent is perpendicular to the radius drawn to that point.
From a point outside a circle, exactly two tangents can be drawn. They are equal in length. The line joining the external point to the centre bisects the angle between the two tangents, and the two tangents subtend equal angles at the centre.
The angle between a tangent and a chord through the point of contact is equal to the angle made by that chord in the alternate segment (the segment on the other side of the chord).
If two chords \(AB\) and \(CD\) intersect inside the circle at \(P\), then \(PA \cdot PB = PC \cdot PD\). If two secants \(PAB\) and \(PCD\) are drawn from an external point \(P\), then \(PA \cdot PB = PC \cdot PD\). If a tangent \(PT\) and a secant \(PAB\) are drawn from \(P\), then \(PT^2 = PA \cdot PB\).
C. Important Formulas and Theorems
D. Step-by-Step Solved Examples
Level 1 Intersecting chords
Two chords \(AB\) and \(CD\) intersect inside a circle at \(P\). If \(PA = 4\text{ cm}\), \(PB = 6\text{ cm}\) and \(PC = 3\text{ cm}\), find \(PD\).
Level 2 Tangent–secant
From an external point \(P\), a tangent \(PT\) touches the circle at \(T\). A secant from \(P\) meets the circle at \(A\) (nearer) and \(B\) (farther). If \(PA = 4\text{ cm}\) and \(AB = 12\text{ cm}\), find \(PT\).
Level 3 Two tangents from a point
From an external point \(P\), tangents \(PA\) and \(PB\) touch a circle with centre \(O\) at \(A\) and \(B\). If \(\angle APB = 60^\circ\), find \(\angle AOB\).
E. Exam-Style Questions
In the given circle, \(RT\) is a tangent touching the circle at \(S\). If \(\angle PST = 30^\circ\) and \(\angle SPQ = 60^\circ\), find \(\angle PSQ\).
Show answer
By the alternate segment theorem, \(\angle PQS = \angle PST = 30^\circ\). In \(\triangle PQS\), \(\angle PSQ = 180^\circ - 60^\circ - 30^\circ = 90^\circ\).
Two chords \(AB\) and \(CD\) intersect at \(P\) inside a circle. If \(PA = 9\text{ cm}\), \(PB = 4\text{ cm}\) and \(PC = 6\text{ cm}\), find \(PD\).
Show answer
\(PA \cdot PB = PC \cdot PD\): \(9 \times 4 = 6 \times PD\), so \(36 = 6\,PD\) and \(PD = 6\text{ cm}\).
The length of a tangent from an external point \(P\) to a circle is \(12\text{ cm}\). A secant from \(P\) cuts the circle at \(A\) and \(B\) with \(PA = 6\text{ cm}\). Find \(AB\).
Show answer
\(PT^2 = PA \cdot PB\): \(144 = 6 \times PB\), so \(PB = 24\text{ cm}\). Thus \(AB = PB - PA = 24 - 6 = 18\text{ cm}\).
From an external point \(P\), two tangents \(PA\) and \(PB\) are drawn to a circle with centre \(O\) and radius \(5\text{ cm}\). If \(OP = 13\text{ cm}\), find the length of each tangent.
Show answer
\(OA \perp PA\), so in right \(\triangle OAP\): \(PA^2 = OP^2 - OA^2 = 169 - 25 = 144\). Hence \(PA = PB = 12\text{ cm}\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Using the tangent–secant formula with only the near part of the secant. | Use the whole secant: \(PT^2 = PA \cdot PB\), where \(PB\) is the distance to the far point. |
| Confusing the intersecting-chords rule with a difference rather than a product. | It is a product of the two segments of each chord: \(PA \cdot PB = PC \cdot PD\). |
| Applying the alternate segment theorem to the wrong segment. | The angles are in the segment on the opposite side of the chord from the tangent. |
| Forgetting that the radius is perpendicular to the tangent. | Always mark the right angle at the point of contact before using Pythagoras. |
G. Exam Tips
- Identify the type of configuration first: two chords, a tangent and a secant, or two tangents.
- For tangent problems, join the point of contact to the centre to create a right angle.
- Write the relevant product relation before substituting, to avoid mixing up segments.
H. Quick Revision Sheet – Tangents and Intersecting Chords
- Tangent \(\perp\) radius at the point of contact.
- Tangents from an external point are equal: \(PA = PB\).
- Alternate segment theorem: tangent–chord angle \(=\) angle in the alternate segment.
- Intersecting chords: \(PA \cdot PB = PC \cdot PD\).
- Tangent–secant: \(PT^2 = PA \cdot PB\).
Topic 5 – Constructions
A. Learning Objectives
After studying this topic you should be able to:
- Construct the circumcircle of a triangle using the perpendicular bisectors of the sides.
- Construct the incircle of a triangle using the angle bisectors.
- Construct tangents to a circle from an external point.
- Construct a triangle similar to a given triangle with a given scale factor.
- Use only a ruler and compasses, and leave all construction arcs in place.
B. Basic Concepts and Definitions
The circle passing through all three vertices of a triangle is the circumcircle. Its centre, the circumcentre, is the point of intersection of the perpendicular bisectors of the sides, and is equidistant from the three vertices.
The circle touching all three sides of a triangle is the incircle. Its centre, the incentre, is the point of intersection of the angle bisectors, and is equidistant from the three sides.
To construct the tangents from an external point \(P\) to a circle with centre \(O\): join \(OP\), draw the perpendicular bisector of \(OP\) to get its midpoint \(M\), then draw the circle with centre \(M\) and radius \(MO\). Where this circle meets the given circle are the points of contact; join \(P\) to those points.
C. Standard Construction Steps
D. Step-by-Step Solved Examples
Level 1 Circumcircle
Construct a triangle \(ABC\) with \(AB = 5\text{ cm}\), \(BC = 6\text{ cm}\) and \(CA = 7\text{ cm}\), and draw its circumcircle.
Level 2 Tangent from an external point
Draw a circle of radius \(3\text{ cm}\). From a point \(P\), \(5\text{ cm}\) from the centre \(O\), construct the two tangents to the circle.
Level 3 Similar triangle with a given scale factor
Construct a triangle similar to a given \(\triangle ABC\) with scale factor \(\dfrac{2}{3}\) (the new triangle is smaller).
E. Exam-Style Questions
Draw a triangle \(ABC\) with sides \(AB = 4\text{ cm}\), \(BC = 5\text{ cm}\) and \(CA = 6\text{ cm}\). Construct its circumcircle and measure the radius.
Show answer
Construct the triangle as described, then draw the perpendicular bisectors of two sides. Their intersection is the circumcentre \(O\); with radius \(OA\) draw the circumcircle. Measuring \(OA\) gives the circumradius (about \(3.0\text{ cm}\) for these sides).
Draw a circle of radius \(4\text{ cm}\). Draw two tangents to the circle from a point \(10\text{ cm}\) from its centre.
Show answer
Mark centre \(O\), draw the circle of radius \(4\text{ cm}\), and mark \(P\) with \(OP = 10\text{ cm}\). Draw the perpendicular bisector of \(OP\) to get its midpoint \(M\). With centre \(M\) and radius \(MO\), draw a circle cutting the given circle at \(A\) and \(B\). Join \(PA\) and \(PB\), the required tangents.
Construct the incircle of a triangle \(ABC\) with \(AB = BC = CA = 6\text{ cm}\), and state the incentre’s position.
Show answer
Construct the equilateral triangle. Bisect any two angles; their intersection \(I\) is the incentre, which is also the centroid and circumcentre for an equilateral triangle. Drop a perpendicular from \(I\) to a side to set the inradius, then draw the incircle.
Construct a triangle similar to a given triangle \(ABC\) with scale factor \(\dfrac{5}{3}\) (the new triangle is larger).
Show answer
Draw a ray through a vertex and mark 5 equal divisions; join the 3rd point to the opposite vertex, then draw a parallel through the 5th point. The triangle formed is similar to \(ABC\) with the required scale factor.
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Using angle bisectors to find the circumcentre. | Circumcentre uses perpendicular bisectors; incentre uses angle bisectors. |
| Not leaving the construction arcs. | Keep all arcs visible; marks are awarded for correct construction lines. |
| Choosing the wrong number of divisions for a similar triangle. | For scale factor \(\frac{m}{n}\), mark \(n\) divisions on the ray and draw the parallel through the \(m\)th point. |
| Free-hand drawing instead of using compasses. | Use ruled lines and compasses throughout; measurements must be accurate. |
G. Exam Tips
- Write down the construction steps briefly before drawing; this earns method marks.
- For the circumcircle, two perpendicular bisectors are enough; the third is a check.
- For tangents from an external point, the auxiliary circle on the diameter \(OP\) is the key step.
- Keep your compass setting unchanged within a step to avoid inaccuracy.
H. Quick Revision Sheet – Constructions
- Circumcircle: perpendicular bisectors meet at the circumcentre.
- Incircle: angle bisectors meet at the incentre.
- Tangents from an external point: circle on \(OP\) as diameter locates the points of contact.
- Similar triangle with scale factor \(\frac{m}{n}\): use \(n\) divisions and draw a parallel through the \(m\)th.
- Leave all arcs; use ruler and compasses only.
Now that you have revised all five topics, attempt the topic quiz for instant feedback, then the practice paper and finally the timed chapter test.