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Lines, angles and triangles for CDS

Parallel lines, the angle-sum and exterior-angle properties, the triangle inequality, Pythagoras, congruence, similarity, the angle bisector theorem and the four centres of a triangle. Six worked CDS-level questions and practice.

10 Oct 2026 7 min read

In this guide
  1. Lines and angles
  2. Triangle basics
  3. Congruence and similarity
  4. The four centres
  5. Worked questions
  6. Practice set
  7. What to do next

The CDS syllabus lists geometry in plain words: lines and angles, the properties of triangles, congruence, similarity, and the concurrence of medians and altitudes. In the paper these become short questions. You get an angle to find, a side to compute, or a ratio of areas. Most need one theorem and two lines of working.

The skill is recognition. Once you see that a figure contains parallel lines, a similar triangle or a centroid, the answer follows quickly. This guide collects the results you need, with the reason behind each, so that you can recognise them under time pressure.

Lines and angles

  • Angles on a straight line add to 180°. Angles around a point add to 360°.
  • Vertically opposite angles are equal.
  • When a transversal cuts two parallel lines:
    • corresponding angles are equal
    • alternate interior angles are equal
    • co-interior angles (same side, between the lines) add to 180°

The converse also holds. If alternate angles are equal, the lines are parallel. Questions often hide the parallel lines inside a larger figure, so look for them first.

Triangle basics

  • Angle sum: the three angles add to 180°.
  • Exterior angle: an exterior angle equals the sum of the two opposite interior angles. Why: both it and the two opposite angles make 180° with the third angle.
  • Sides and angles: the largest side is opposite the largest angle. Equal sides face equal angles.
  • Triangle inequality: any two sides together are longer than the third. So the third side lies strictly between the difference and the sum of the other two.

Right, acute or obtuse

Let c be the longest side. Compare c² with a² + b²:

ComparisonTriangle
c² = a² + b²Right-angled (Pythagoras)
c² < a² + b²Acute-angled
c² > a² + b²Obtuse-angled

Pythagorean triplets save time: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29 and 9-40-41, plus their multiples (6-8-10, 9-12-15 and so on).

Right triangle facts

  • The median to the hypotenuse is half the hypotenuse, because the midpoint of the hypotenuse is the centre of the circle through all three vertices.
  • The altitude to the hypotenuse = (product of legs) ÷ hypotenuse. Both expressions are twice the area divided by a base.

Congruence and similarity

Congruent triangles are identical in shape and size. The tests are SSS, SAS, ASA, AAS and, for right triangles, RHS. SSA and AAA are not tests of congruence.

Similar triangles have the same shape. The tests are AA, SSS (sides in proportion) and SAS (two sides in proportion with the included angle equal). For similar triangles with sides in the ratio k:

  • perimeters, medians, altitudes and angle bisectors are also in the ratio k
  • areas are in the ratio k²

Three theorems built on similarity

  • Basic proportionality theorem: a line parallel to one side of a triangle divides the other two sides in the same ratio. If DE ∥ BC, then AD/DB = AE/EC.
  • Midpoint theorem: the segment joining the midpoints of two sides is parallel to the third side and half its length.
  • Angle bisector theorem: the bisector of angle A meets BC at D so that BD : DC = AB : AC.

The four centres

CentreWhere lines meetKey property
Centroid (G)MediansDivides each median 2 : 1 from the vertex
Incentre (I)Angle bisectorsEquidistant from the sides; ∠BIC = 90° + A/2
Circumcentre (O)Perpendicular bisectors of sidesEquidistant from vertices; ∠BOC = 2A if A is acute
Orthocentre (H)Altitudes∠BHC = 180° − A in an acute triangle

In a right triangle the circumcentre is the midpoint of the hypotenuse and the orthocentre is the right-angle vertex. In an equilateral triangle all four centres coincide.

For an equilateral triangle of side a: height = (√3/2)a, circumradius = a/√3 (two-thirds of the height), and inradius = a/(2√3) (one-third of the height).

Worked questions

Question 1: Two parallel lines are cut by a transversal. The two co-interior angles on one side are (3x + 10)° and (2x + 20)°. Find them.

  • Co-interior angles add to 180°: 5x + 30 = 180, so x = 30.
  • The angles are 100° and 80°.

Question 2: In triangle ABC, ∠B = 60° and ∠C = 40°. The bisectors of angles B and C meet at I. Find ∠BIC.

  • ∠A = 180° − 100° = 80°, so ∠BIC = 90° + 40° = 130°.
  • Check directly: in triangle BIC the angles at B and C are 30° and 20°, so ∠BIC = 130°.

Question 3: Two sides of a triangle are 5 cm and 9 cm. How many integer values can the third side take?

  • 9 − 5 < x < 9 + 5, so 4 < x < 14.
  • x = 5, 6, …, 13, which is 9 values.

Question 4: In triangle ABC, D is on AB and E on AC with DE ∥ BC. AD = 4 cm, DB = 6 cm and AE = 6 cm. Find EC and the ratio of the area of triangle ADE to that of ABC.

  • By the basic proportionality theorem, 4/6 = 6/EC, so EC = 9 cm.
  • Triangles ADE and ABC are similar with ratio AD : AB = 4 : 10. Areas are in the ratio 4 : 25.

Question 5: In triangle ABC, AB = 12 cm, AC = 8 cm and BC = 15 cm. The bisector of angle A meets BC at D. Find BD.

  • BD : DC = 12 : 8 = 3 : 2.
  • BD = 15 × 3/5 = 9 cm, and DC = 6 cm.

Question 6: A right triangle has legs 6 cm and 8 cm. Find the altitude to the hypotenuse and the median to the hypotenuse.

  • Hypotenuse = 10 cm.
  • Altitude = 6 × 8 ÷ 10 = 4.8 cm. Median = half the hypotenuse = 5 cm.

Practice set

  1. The angles of a triangle are x, 2x and 3x. Find them.
  2. The legs of a right triangle are 7 cm and 24 cm. Find the hypotenuse.
  3. The areas of two similar triangles are in the ratio 16 : 25. Find the ratio of their corresponding sides.
  4. An exterior angle of a triangle is 120°, and the two opposite interior angles are in the ratio 1 : 3. Find all three angles.
  5. Is a triangle with sides 5, 7 and 9 acute, right or obtuse?
  6. In an acute triangle ABC, ∠A = 50° and H is the orthocentre. Find ∠BHC.
  7. A median of a triangle is 15 cm long. How far is the centroid from the midpoint of the side?
  8. The height of an equilateral triangle is 9 cm. Find its circumradius.

Answers

  1. 30°, 60°, 90°. 6x = 180.
  2. 25 cm. It is the 7-24-25 triplet.
  3. 4 : 5. Take square roots.
  4. 30°, 90° and 60°. The opposite angles share 120° as 30° and 90°. The third angle is 180° − 120° = 60°.
  5. Obtuse. 9² = 81 is greater than 5² + 7² = 74.
  6. 130°. 180° − 50°.
  7. 5 cm. The centroid is 10 cm from the vertex and 5 cm from the midpoint.
  8. 6 cm. The circumradius is two-thirds of the height.

What to do next

  • Draw each of the four centres once in an acute triangle, and label its angle formula
  • Learn the six Pythagorean triplets and their first two multiples
  • Continue with circles, which reuses the angle and similarity facts
  • Then revise quadrilaterals and polygons
  • Solve the triangle questions from the last five CDS papers, noting which theorem each one used

A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Union Public Service Commission website .

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