Congruence and Similarity of Triangles

Table of Contents
1. Congruence and similarity of triangles for SSC
2. Conditions for Congruence of Triangles
3. Conditions for Similarity of Triangles
4. Consequences and useful relations for similar triangles
5. Congruence and similarity of triangles for SSC: Important Theorems
6. Applications, problem-solving tips and quick checks
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Geometry is a crucial subject in SSC Quantitative Aptitude, focusing on shapes and sizes. To excel in this area, one must master the fundamentals, theorems, and practical applications. This document explains congruence and similarity of triangles, the key ideas used frequently in competitive quantitative problems.

Congruence and similarity of triangles for SSC

Two triangles are congruent when they are identical in both shape and size. This means that their corresponding sides and corresponding angles are equal.

Two triangles are similar when they have the same shape but possibly different sizes. This means their corresponding angles are equal and corresponding sides are in proportion.

CPCTC: Corresponding Parts of Congruent Triangles are Congruent.

Conditions for Congruence of Triangles

SSS (Side-Side-Side)

All three pairs of corresponding sides are equal. If in △ABC and △DEF,
AB = DE, BC = EF, CA = FD,
then △ABC ≅ △DEF.

SAS (Side-Angle-Side)

Two pairs of corresponding sides and the included angle (the angle between those sides) are equal. If in △ABC and △DEF,
AB = DE, AC = DF and ∠A = ∠D,
then △ABC ≅ △DEF.

AAS (Angle-Angle-Side)

Two pairs of corresponding angles and a corresponding non-included side are equal. If in △ABC and △DEF,
∠A = ∠D, ∠C = ∠F and AB = DE,
then △ABC ≅ △DEF.

ASA (Angle-Side-Angle)

Two pairs of corresponding angles and the included side (the side between those angles) are equal. If in △ABC and △DEF,
∠A = ∠D, ∠C = ∠F and AC = DF,
then △ABC ≅ △DEF.

RHS (Right-Hypotenuse-Side)

For right triangles, if the hypotenuse and one corresponding side are equal, the triangles are congruent. If in right triangles △ABC and △DEF,
∠B = ∠E = 90°, AC = DF and BC = EF,
then △ABC ≅ △DEF.

Conditions for Similarity of Triangles

AA (Angle-Angle)

Two pairs of corresponding angles are equal. If △ABC and △FDE have two corresponding equal angles, then the triangles are similar: △ABC ∼ △FDE.

SAS (Side-Angle-Side) for similarity

Two pairs of corresponding sides are in the same ratio and the included angles are equal. If in △ABC and △DEF,
AB/DE = BC/EF and ∠B = ∠E,
then △ABC ∼ △DEF.

SSS (Side-Side-Side) for similarity

All three pairs of corresponding sides are proportional. If
AB/DE = BC/EF = CA/FD,
then △ABC ∼ △DEF.

Consequences and useful relations for similar triangles

• Ratio of areas of two similar triangles = square of ratio of corresponding sides.
• Ratio of corresponding sides = ratio of corresponding altitudes (heights).
• Ratio of corresponding sides = ratio of corresponding medians.
• Ratio of corresponding sides = ratio of corresponding internal angle bisectors.
• Ratio of corresponding sides = ratio of in-radii (and also circum-radii) and equals ratio of perimeters.

The above tests and relations form the backbone for triangle problems. Mastering them, together with practice, gives quick answers under time pressure.

Congruence and similarity of triangles for SSC: Important Theorems

Basic Proportionality Theorem (Thales' theorem)

If a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. For triangle ABC with a line DE parallel to BC meeting AB at D and AC at E,

If DE ∥ BC then AD/DB = AE/EC.

Example: Area of quadrilateral DECB is 180 cm² and DE divides AC in the ratio 2:5. If DE ∥ BC, what is the area of △ADE?

a) 16 cm²
b) 32 cm²
c) 40 cm²
d) 20 cm²

Answer: a
Solution: Since DE divides AC in the ratio 2:5,
\( \dfrac{AE}{EC} = \dfrac{2}{5} \)
By the Basic Proportionality Theorem,
\( \dfrac{AE}{EC} = \dfrac{AD}{DB} \)
Therefore, AD is a fraction of AB. Express AD/AB in terms of AE and EC.
\( \dfrac{AD}{AB} = \dfrac{AE}{AE + EC} \)
\( \dfrac{AD}{AB} = \dfrac{2}{2 + 5} = \dfrac{2}{7} \)
Hence, triangles △ADE and △ABC are similar (by SAS similarity: corresponding sides in ratio and included angle equal because DE ∥ BC implies corresponding angles equal).
Ratio of their areas = square of ratio of corresponding sides.
\( \dfrac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \left(\dfrac{2}{7}\right)^2 = \dfrac{4}{49} \)
Let area of △ADE = 4x. Then area of △ABC = 49x.
Total area relation: area of △ABC = area of △ADE + area of quadrilateral DCEB.
\( 49x = 4x + 180 \)
\( 45x = 180 \)
\( x = 4 \)
Area of △ADE = 4x = 16 cm².

Angle Bisector Theorem

If a bisector of an angle of a triangle divides the opposite side, it divides that side in the ratio of the adjacent sides. In triangle ABC, if AD is the bisector of ∠A and meets BC at D, then

BD/DC = AB/AC.

Example: If AD bisects ∠BAC and AD divides BC in the ratio 1:1 then, the ratio of area of △ABC to the area of △ABD is:

a) 3:1
b) 4:1
c) 3:2
d) 2:1

Answer : d
Solution: Given that AD bisects ∠A and AD divides BC in the ratio 1:1, therefore
\( \dfrac{BD}{DC} = \dfrac{AB}{AC} = 1 \)
Thus AB = AC and BD = DC.
AD = AD (common side). Therefore the two triangles △ABD and △ACD have three pairs of equal sides:
AB = AC, BD = DC, AD = AD.
Hence △ABD ≅ △ACD (by SSS congruence).
Area of △ABC = Area of △ABD + Area of △ACD = 2 × Area of △ABD.
Required ratio = Area(△ABC) : Area(△ABD) = 2 : 1.

Applications, problem-solving tips and quick checks

• Always identify corresponding vertices before comparing sides or angles. Write triangles in the same order (e.g., △ABC and △DEF) so that corresponding parts are clear.
• For congruence use SSS, SAS, ASA, AAS and RHS. For similarity use AA, SAS (similarity), SSS (similarity).
• When a line is parallel to a triangle side, use the Basic Proportionality Theorem to convert lengths into ratios and then apply similarity to get areas or lengths.
• For area problems, remember area ratios follow the square of linear ratios. Use this to scale areas quickly.
• Angle bisector problems often reduce to ratio relations between sides; check for congruence after converting ratios.
• Label diagrams clearly. If a numeric ratio is given for a side division, convert it into fractional parts of the whole side (for example 2:5 makes parts 2/7 and 5/7 of the whole).

Understanding these rules, practicing many problems, and checking each step for correct correspondence will ensure accuracy and speed in exams and competitive tests.