Master the Similarity of Triangles: Detailed Concepts, Proofs, and Theorems

In geometry, triangles are similar when they share the same shape, irrespective of their size. This happens when:

Corresponding angles are equal.
Corresponding sides are in the same ratio (proportion).

For example, if $\boldsymbol{\triangle ABC \sim \triangle DEF}$ , then:

$\boldsymbol{\angle A = \angle D, \, \angle B = \angle E, \, \angle C = \angle F}$ $\displaystyle\boldsymbol{\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}}$

This concept is foundational in geometry and has practical applications in fields such as construction, navigation, and design.

Basic Proportionality Theorem (Thales’ Theorem)

Statement

If a line is drawn parallel to one side of a triangle, intersecting the other two sides, the line divides those sides in the same ratio.

In $\boldsymbol{\triangle ABC}$ , if $\boldsymbol{DE \parallel BC}$ , then:

$\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$

Understanding Thales’ Theorem

Setup

Draw any angle $\boldsymbol{\angle XAY}$ and on one arm $\boldsymbol{AX}$ , mark points $\boldsymbol{P, Q, D, R, B}$ such that: $\boldsymbol{AP = PQ = QD = DR = RB}$
Draw another arm $\boldsymbol{AY}$ and mark a point $\boldsymbol{C}$ on it.
From $\boldsymbol{D}$ , draw $\boldsymbol{DE \parallel BC}$ .

Observations

Measure $\boldsymbol{AD}$ , $\boldsymbol{DB}$ , $\boldsymbol{AE}$ , and $\boldsymbol{EC}$ .
You will find:

$\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$

This supports the Basic Proportionality Theorem.

Proof of Basic Proportionality Theorem

Given

In $\boldsymbol{\triangle ABC}$ , $\boldsymbol{DE \parallel BC}$ intersects $\boldsymbol{AB}$ and $\boldsymbol{AC}$ at $\boldsymbol{D}$ and $\boldsymbol{E}$ .

To Prove

$\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$

Construction

Draw $\boldsymbol{DM \perp AB}$ and $\boldsymbol{EN \perp AC}$ .

Proof

Area of $\boldsymbol{\triangle ADE}$ is proportional to its base $\boldsymbol{AD}$ :
$\displaystyle\boldsymbol{\text{Area of } \triangle ADE = \frac{1}{2} \cdot AD \cdot DM}$
Similarly, for $\boldsymbol{\triangle BDE}$ :
$\displaystyle\boldsymbol{\text{Area of } \triangle BDE = \frac{1}{2} \cdot DB \cdot DM}$
Taking the ratio of areas:
Area of $\displaystyle\boldsymbol{\frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle BDE} = \frac{AD}{DB}}$
Similarly, for $\boldsymbol{\triangle ADE}$ and $\boldsymbol{\triangle DEC}$ :
Area of $\displaystyle\boldsymbol{\frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle DEC} = \frac{AE}{EC}}$
Thus, $\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$ .

Converse of Basic Proportionality Theorem

Statement

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Proof

Suppose $\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$ .
Assume $\boldsymbol{DE}$ is not parallel to $\boldsymbol{BC}$ .
Draw $\boldsymbol{DE'} \parallel BC}$ , and by the Basic Proportionality Theorem: $\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE'}{E'C}}$
But this contradicts $\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$ . Hence, $\boldsymbol{DE \parallel BC}$ .

Examples Based on Theorems

Example 1: Prove $\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC} = \frac{AB}{AC}}$

Solution:
From Thales' Theorem:

$\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$

Adding 1 to both sides:

$\displaystyle\boldsymbol{\frac{AD + DB}{DB} = \frac{AE + EC}{EC}}$ $\displaystyle\boldsymbol{\frac{AB}{DB} = \frac{AC}{EC}}$

Thus:

$\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC} = \frac{AB}{AC}}$

Example 2: Prove $\displaystyle\boldsymbol{\frac{AE}{ED} = \frac{BF}{FC}}$ in a trapezium.

Solution:
Join $\boldsymbol{AC}$ , intersecting $\boldsymbol{EF}$ at $\boldsymbol{G}$ .

Using Thales' Theorem in $\boldsymbol{\triangle AED}$ : $\displaystyle\boldsymbol{\frac{AE}{ED} = \frac{AG}{GC}}$
Similarly, in $\boldsymbol{\triangle BFC}$ : $\displaystyle\boldsymbol{\frac{BF}{FC} = \frac{AG}{GC}}$

Thus:

$\displaystyle\boldsymbol{\frac{AE}{ED} = \frac{BF}{FC}}$

Example 3: Prove $\boldsymbol{\triangle PST \sim \triangle PQR}$ .

Solution:

$\boldsymbol{ST \parallel QR}$ , so by Thales' Theorem: $\displaystyle\boldsymbol{\frac{PS}{PQ} = \frac{PT}{PR}}$
$\boldsymbol{\angle PST = \angle PQR}$ (corresponding angles).

Thus, $\boldsymbol{\triangle PST \sim \triangle PQR}$ .

Practice Problems

In $\boldsymbol{\triangle ABC}$ , prove that $\displaystyle\boldsymbol{\frac{AD}{DB} = \frac{AE}{EC}}$ when $\boldsymbol{DE \parallel BC}$ .
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
In $\boldsymbol{\triangle PQR}$ , $\boldsymbol{ST \parallel QR}$ . If $\boldsymbol{PS = 5}$ cm, $\boldsymbol{PQ = 10}$ cm, and $\boldsymbol{PR = 15}$ cm, find $\boldsymbol{PT}$ .