PYQ Vault

CDS Mathematics · Triangles

Congruence and Similarity

Congruent triangles match in every side and angle; similar triangles match in angles, so every length of one is the same multiple of the matching length of the other.

Why this matters

Thirteen PYQs. The congruence questions test which rule applies and why equal angles alone are not enough. The similarity questions turn on one pattern — a shared angle plus one more equal angle — and on reading the vertex order in a statement like △ABR ∼ △PQR. Two recent items ask where two crossing lines meet between two poles, which has a one-line answer.

Concept 1 of 3: The congruence rules

Three pieces of a triangle fix the whole triangle only if they are the right three. Three sides do; two sides with the angle between them do; two angles with a side do. Three angles never do, because they fix the shape but not the size.

Definition

Two triangles are congruent by:

  • SSS — three sides equal;
  • SAS — two sides and the angle between them;
  • ASA / AAS — two angles and any matching side;
  • RHS — right angle, hypotenuse and one other side.

AAA gives only similarity. SSA (the angle not between the sides) is not a rule, except for the right-angled case RHS.

Rules that fix a triangle

SSS, SAS, ASA, AAS, RHS(not AAA, not SSA)\text{SSS},\ \text{SAS},\ \text{ASA},\ \text{AAS},\ \text{RHS} \qquad (\text{not AAA, not SSA})

Worked example

In quadrilateral ABCDABCD, AB=ADAB = AD and CB=CDCB = CD. Show that ACAC bisects ∠A\angle A.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q98Moderate

Example 1 · Triangles · Congruence and Similarity

In the figure given below, M is the mid-point of AB and ∠DAB=∠CBA\angle DAB = \angle CBA and ∠AMC=∠BMD\angle AMC = \angle BMD. Then the triangle ADM is congruent to the triangle BCM by

The angle must be the included one

SAS needs the angle between the two named sides. A statement listing 'two sides and an angle' without saying which angle is not a valid rule.

Concept 2 of 3: Similar triangles and the shared-angle pattern

Two equal angles force the third to be equal as well, so two angles are enough for similarity. Then one triangle is a scaled copy of the other: every side, perimeter, altitude and median grows by the same factor.

Definition

  • AA: two pairs of equal angles make triangles similar.
  • If the scale factor is kk, every corresponding length (side, perimeter, altitude, median, bisector) is in the ratio kk.
  • Shared angle: if DD is on BCBC and ∠ADC=∠BAC\angle ADC = \angle BAC, then triangles ADCADC and BACBAC share ∠C\angle C, so they are similar and DCAC=ACBC\dfrac{DC}{AC} = \dfrac{AC}{BC}, i.e. AC2=DC⋅BCAC^2 = DC\cdot BC.
  • In a statement like △ABR∼△PQR\triangle ABR \sim \triangle PQR, the vertices match in the order written: AA with PP, BB with QQ, RR with RR.

Shared-angle similarity

∠ADC=∠BAC  ⇒  AC2=DC⋅BC\angle ADC = \angle BAC \;\Rightarrow\; AC^2 = DC \cdot BC

Worked example

DD is on BCBC with ∠CAD=∠B\angle CAD = \angle B. If BC=16BC = 16 and CD=4CD = 4, find CACA.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q84Moderate

Example 2 · Triangles · Congruence and Similarity

ABC is a triangle and D is a point on the side BC. If BC = 12 cm, BD = 9 cm and ∠ADC=∠BAC\angle ADC = \angle BAC, then the length of AC is equal to

Match vertices by the statement, not by the letters' positions

In △ABR∼△PQR\triangle ABR \sim \triangle PQR, BRBR matches QRQR and ARAR matches PRPR. Write the three pairs out before dividing, or the scale factor gets applied to the wrong side.

Concept 3 of 3: Crossing lines between two poles

Join the top of each pole to the foot of the other. The two lines cross at a height that depends only on the two pole heights, not on how far apart the poles stand — move them apart and the crossing point moves, but its height does not.

Definition

Poles of heights aa and bb stand a distance dd apart; the crossing point is at height hh and at distance xx from the foot of the first pole.

  • Similar triangles with the second pole: hb=xd\dfrac hb = \dfrac xd.
  • Similar triangles with the first pole: ha=d−xd\dfrac ha = \dfrac{d - x}{d}.
  • Adding: ha+hb=1\dfrac ha + \dfrac hb = 1, so h=aba+bh = \dfrac{ab}{a + b}, and dd has cancelled.

Height of the crossing point

h=aba+bh = \dfrac{ab}{a + b}

Worked example

Poles of heights 66 m and 1212 m stand 2020 m apart. Lines join the top of each to the foot of the other. How high above the ground do they cross?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (II) 2025 — Elementary Mathematics · Q66Moderate

Example 3 · Triangles · Congruence and Similarity

Two poles of heights 10 m and 15 m are 25 m apart. What is the height of the point of intersection of the lines joining the tip of each pole to the foot of the other pole?

Not the average

Poles of 1212 and 2424 give 88, not 1818 (the average) and not anything that uses the distance between them. The crossing height is always less than the shorter pole.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The congruence rules

    Rules that fix a triangle

    SSS, SAS, ASA, AAS, RHS(not AAA, not SSA)\text{SSS},\ \text{SAS},\ \text{ASA},\ \text{AAS},\ \text{RHS} \qquad (\text{not AAA, not SSA})
  • Similar triangles and the shared-angle pattern

    Shared-angle similarity

    ∠ADC=∠BAC  ⇒  AC2=DC⋅BC\angle ADC = \angle BAC \;\Rightarrow\; AC^2 = DC \cdot BC
  • Crossing lines between two poles

    Height of the crossing point

    h=aba+bh = \dfrac{ab}{a + b}

Watch out for (3)

Test yourself on Triangles

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.