Similarity of Triangles: Formula, Theorems, Properties and Examples

Triangles are one of the many basic shapes that are used in geometry. Basically, polygons having three sides and three vertices can be defined as a triangle. Earlier we proved congruency between two triangles. However, one might wonder, is a similarity between two triangles different from congruency? Well, yes! Congruency is when the triangles have similar sizes as well as similar shapes. The similarity between any two triangles is dependent on the concept of the same shapes and is one of the most vital phenomena in Geometry that acts as a foundation to the advanced Triangle theorems or concepts. 

There are two conditions for proving the similarities of two triangles. These are:

• When the corresponding angles of both triangles are equal.

• When the corresponding sides of both the triangles are also in the same ratio.

For eg, if there are two triangles ΔABC and ΔEFG, and

• ∠A = ∠E, ∠B= ∠F, and ∠C= ∠G. 
• Also, if AB/EF, AC/EG, and BC/FG. 

This implies that both the  Δ ABC and Δ EFG are similar to each other.

---

Properties of Similar Triangles

[Click Here for Sample Questions]

There are basically three properties on the basis of which two triangles of different sizes can be proven similar:

• The shape of the two triangles is similar but the size of the two triangles is not similar.
• The corresponding angles of both the triangles are equal.
• The ratio of sides of both triangles is equal.

We can say that any two triangles are similar if-

1. They have congruent corresponding angles.
2. They have proportional corresponding sides.

---

Theorems based on Similar Triangles

[Click Here for Sample Questions]

There are three theorems used to deal with problems based on similar triangles. These are:

• AA(Angle-Angle) Similarity 

According to this theorem, if two angles of the given triangles are equal then those two triangles are considered to be similar.

In the triangles ABC and EFG, if ∠A=∠E and ∠B=∠F 

then, in that case, both the triangles ?ABC and ?EFG are similar to each other.

ΔABC ~ΔEFG.

• SSS(Side-Side-Side) Similarity

SAS theorem implies that the two triangles are considered to be similar if all three sides of both the triangles are in proportion to each other 

For eg, in ΔXYZ and ΔEFG

if XY/EF = YZ/FG = XZ/EG then both the triangles are said to be similar to each other.

• SAS(Side Angle Side) Similarity

According to the SSS similarity theorem, if the two sides of a triangle are in similar proportion to the sides of the other triangle and the angle formed by these two sides is equal, then, the triangles are considered to be similar to each other.

In ΔXYZ and ΔEFG, if all the sides of the triangles have the same proportion then it is considered to be similar to each other. 

If XY/EF = YZ/FG and ∠X = ∠E 

Then ΔABC ~ΔEFG

---

Basic Proportionality Theorem

[Click Here for Sample Questions]

The basic proportionality theorem states that, if a line is drawn, parallel to a side of the triangle, to intersect the rest two sides of the triangle at a certain specific point, then the rest of the two sides of the triangle are divided in the same ratio. 

Let’s demonstrate the application of the basic proportionality theorem with the help of an example:

Given: ABC is a triangle in which DE||BC, intersects, AB and AC at D and E respectively.

To prove: AD/DB = AE/EC.

Construction: Join B to E and D to C and draw DM⊥AE and EN⊥ AD.

Proof: ABC is a triangle and DE is a line parallel to BC intersecting the two sides of the triangle i.e. AB and AC at D and E respectively. 

According to the Area of Triangle formula, area = ½ x base x height

Therefore, the area of ΔADE = ½ X AD X EN

Similarly, Δ BDE = ½ X DB X EN

Hence, Δ ADE / Δ BDE = ½ X AD X EN / ½ X DB X EN

= AD/DB

Simialry, Δ ADE/ Δ DEC = ½ X AE X DM / ½ X EC X DM

= AE/EC

In accordance with the property of triangles, the triangles which are drawn on the same base and in between the same parallel lines have equal areas

Therefore, since Δ BDE and Δ DEC are on the same base i.e. DE 

The area of Δ BDE = Δ DEC

Therefore, AD/DB = AE/EC