Construction of Similar Triangles

Similar Triangles

If Two triangles ∆ABC and ∆PQR are said to be similar, following two conditions are satisfied:

1. The corresponding angles of the triangles are equal.


∠A = ∠P, ∠B = ∠Q, ∠C = ∠R


2. Since,∆ABC and ∆PQR are two similar triangles, their corresponding sides are in a ratio or proportion.

That is,

\( \frac {AB}{PQ} \) = \( \frac {BC}{QR} \) = \( \frac {AC}{PR} \)

We write ∆ABC and ∆PQR are similar as ∆ABC~∆PQR

For example;

Consider the two triangles given in the figure,

If ∆ABC~∆PQR, What is length of the side PR if AB = 6 cm, AC = 8 cm and PQ = 3 cm?

Since, ∆ABC~∆PQR

\( \frac {AB}{PQ} \) = \( \frac {AC}{PR} \)

\( \frac {6}{3} \) = \( \frac {8}{PR} \)

PR = \( \frac {8~×~3}{6} \) = 4 cm

Let’s see how to construct similar triangles.

Consider ∆ABC where BC = 6 cm,∠B=40°and ∠C=60°. Draw a triangle similar to ∆ABC with a scale factor 2.

Scale factor is the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

Figure 1-Triangle ABC

Here, a scale factor of 2 means that sides of the new triangle which is similar to ∆ABC are twice the sides of ∆ABC.

Let ∆PQR be the new triangle.

QR = 2 × 6 = 12 cm [Scale factor is 2]

∠B = ∠Q = 40°and ∠C = ∠R=60°.

  • Draw QR of length 12 cm
  • Draw a line through B which makes an angle of 40° from BC.
  • Draw a line through C which makes an angle of 60° from BC.
  • Mark the intersection point of above two lines as P. ∆PQR is the required triangle (Refer figure).

Figure 2-How to construct similar triangles

Now, suppose the scale factor is a fraction,like \( \frac 54 \), \( \frac 78 \) etc or suppose we don’t know length of the sides?

Then we won’t be able to construct similar triangles precisely.

The method to construct a similar triangle precisely is discussed here.

Problem: Construct a triangle which is similar to ∆ABC with scale factor \( \frac 35 \).

Scale factor \( \frac 35 \) means, the new triangle will have side lengths \( \frac 35 \) times the corresponding side lengths.

Construction of Similar Triangles

The steps of construction of similar triangles are as follows (Refer figure)

  • Draw a ray BX which makes acute angle with BC on the opposite side of vertex A.
  • Locate 5 points on the ray BX and mark them as B1, B2, B3, B4 and B5 such that B B1 = B1 B2 = B2 B3 = B4 B5.
  • Join B5C
  • Draw a line parallel to B5C through B3 [Since 3 is the smallest among 3 and 5] and mark C’ where it intersects with BC.
  • Draw a line through the point C’ parallel to AC and mark A’ where it intersects AB.
  • A’BC’ is the required triangle.

Figure 3-How to construct similar triangles

How can we verify that ∆ABC~∆A’BC’ ?

\( \frac {BC’}{C’C}\) = \( \frac 32 \) [By construction]


\( \frac {BC}{BC’}\) = \( \frac {BC’~+~C’C}{BC’}\) = \( 1~+~\frac 23 \) = \( \frac 53 \)

That gives,

\( \frac {BC’}{BC}\) = \( \frac 35 \)<

And, since A’C’ is parallel to AC,

∠ A’C’B=∠ACB [corresponding angles]

∠ABC=∠A’BC’ [common angle]


∆ABC ~ ∆A’BC’