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CAT 2025 Lesson : Lines & Triangles - Congruency & Similarity

3.9 Congruency of Triangles

Congruent triangles are identical triangles whose corresponding sides and corresponding angles are equal. In other words, when one triangle is placed over the other, the vertices and sides will coincide.

As congruent triangles are identical triangles, all of their corresponding parameters such as area, perimeter, circumradius, inradius, altitudes, bisectors, etc. will be equal.

Naming Convention: The symbol ≅ is used to indicate congruence. The triangles are named in the same order in which the vertices will overlap one another when placed one over the other. For instance, if

\triangle

ABC ≅

\triangle

DEF, then from the corresponding letter positions we note that AB

=

DE, BC

=

EF, CA

=

FD,

\angle

=

\angle

\angle

=

\angle

E and

\angle

=

\angle

F.

Two triangles are said to be congruent if the following rules are met.

Rules	Figure
Side-Side-Side (SSS Rule): When the corresponding sides of two triangles are equal in length, then the two triangles are congruent. If AB $=$ DE, BC $=$ EF and CA $=$ FE, then $\triangle$ ABC ≅ $\triangle$ DEF
Side-Angle-Side (SAS Rule): In two triangles, if $2$ pairs of corresponding sides are equal and the angle subtended between them are equal, then the two triangles are congruent. If AB $=$ DE, BC $=$ EF and $\angle$ ABC $=$ $\angle$ DEF, then $\triangle$ ABC ≅ $\triangle$ DEF
Angle-Side-Angle (ASA Rule): In two triangles, if $2$ pairs of angles are equal and the corresponding sides common to both the angles are equal, then the two triangles are congruent. If $\angle$ ABC $=$ $\angle$ DEF, $\angle$ BCA $=$ $\angle$ EFD and BC $=$ EF, then $\triangle$ ABC ≅ $\triangle$ DEF
Right Angle-Hypotenuse-Side (RHS Rule): In two right-angled triangles, if the length of the hypotenuses are equal and that of one of the other two sides are equal, then the two triangles are congruent. If $\angle$ ABC $=$ $\angle$ DEF $=$ $90^\mathrm{o}$ , AC $=$ DF and AB $=$ DE, then $\triangle$ ABC ≅ $\triangle$ DEF

3.10 Similarity of Triangles

Two triangles are said to be similar if their corresponding angles are equal. In similar triangles, the corresponding sides will be in the same proportion. ~ Is the symbol used to denote similarity. The naming convention stated for congruence of triangles applies here as well.

Rules	Figure
Side-Side-Side (SSS Rule): When the corresponding sides of two triangles are in the same proportion, then the two triangles are similar. If $\dfrac{ \mathrm{AB}}{ \mathrm{DE}}$ $=$ $\dfrac{ \mathrm{BC}}{ \mathrm{EF}}$ $=$ $\dfrac{ \mathrm{CA}}{ \mathrm{FD}}$ , then $\triangle$ ABC ~ $\triangle$ DEF
Side-Angle-Side (SAS Rule): In two triangles, if $2$ pairs of corresponding sides are in the same proportion and the angle subtended between them are equal, then the two triangles are similar. If $\dfrac{\mathrm{AB}}{\mathrm{DE}} = \dfrac{\mathrm{BC}}{\mathrm{EF}}$ , and $\angle$ ABC = $\angle$ DEF, then $\triangle$ ABC ~ $\triangle$ DEF
Angle-Angle (AA Rule): In two triangles, if $2$ pairs of angles are equal, then the third should also be equal (as sum of angles is $180^\mathrm{o}$ . Therefore, if $2$ pairs of angles are equal in $2$ triangles, then they are similar. If $\angle$ ABC $=$ $\angle$ DEF and $\angle$ BCA $=$ $\angle$ EFD, then $\triangle$ ABC ~ $\triangle$ DEF

When

\triangle

ABC ~

\triangle

DEF, the following are to be noted.

1)

\angle

=

\angle

\angle

=

\angle

\angle

=

\angle

F

2) Ratio of corresponding sides are equal,
i.e.

\dfrac{ \mathrm{AB}}{ \mathrm{DE}} = \dfrac{ \mathrm{BC}}{ \mathrm{EF}} = \dfrac{ \mathrm{CA}}{ \mathrm{FD}}

3) Ratio of Sides

=

Ratio of Perimeters

=

Ratio of Altitudes

=

Ratio of Medians

=

Ratio of Angle Bisectors

=

Ratio of Perpendicular Bisectors

=

Ratio of Inradii

=

Ratio of Circumradii

4) Ratio of areas

=

ratio of squares of corresponding sides,
i.e.

\dfrac{ \mathrm{area}(\triangle \mathrm{ABC})}{ \mathrm{area}(\triangle \mathrm{DEF})}

=

\dfrac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}

=

\dfrac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}} = \dfrac{\mathrm{CA}^{2}}{\mathrm{FD}^{2}}

Example 21

\triangle

ABC, BC

=

6

cm. Point D lies on AB such that AD

=

5

cm and

\angle

BAC

=

\angle

BCD. Then, BD

=

Solution

Let BD

=

x

\triangle

BAC and

\triangle

BCD,

\angle

BAC

=

\angle

BCD (Given in the question)

\angle

CBA

=

\angle

DBC (Common Angle)

\triangle

BAC ~

\triangle

BAC (AA Rule)

\therefore

\dfrac{\mathrm{BA}}{\mathrm{BC}} = \dfrac{\mathrm{BC}}{\mathrm{BD}}

⇒

\dfrac{x + 5}{6} = \dfrac{6}{x}

⇒

x (x + 5) = 36 = 4 \times (4 + 5)

x = 4

Answer:

4

Example 22

In the following

\triangle

ABC, BC | | DE and CD | | EF. If AE

=

8

cm, EC

=

12

cm and AF

=

4

cm, then BD

=

Solution

In a triangle, if a line parallel to the a side is drawn, then the other

24

sides serve as the transversal. As corresponding angles are equal, the inner triangle formed is similar to the outer triangle (AA Similarity rule).

As CD | | EF,

\triangle

AFE ~

\triangle

ADC.

\dfrac{\mathrm{AF}}{\mathrm{AD}} = \dfrac{\mathrm{AE}}{\mathrm{AC}}

⇒

\dfrac{4}{4 + \mathrm{FD}} = \dfrac{8}{8 + 12}

⇒ FD

=

6

As BC | | DE,

\triangle

ADE ~

\triangle

ABC

\dfrac{\mathrm{AD}}{\mathrm{AB}} = \dfrac{\mathrm{AE}}{\mathrm{AC}}

⇒

\dfrac{10}{10 + \mathrm{BD}} = \dfrac{8}{20}

⇒ BD

=

15

Answer:

15

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