Construction : To construct a triangle similar to a given triangle as per given scale factor. This construction involves two different situations. In one, the triangle to be constructed is smaller and in the other it is larger than the given triangle. Here, the scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle (see also Chapter 6). Let us take the following examples for understanding the constructions involved. The same methods would apply for the general case also.

Example: Construct a triangle similar to a given triangle ABC with its sides equal to $3/4$ of the corresponding sides of the triangle ABC (i.e., of scale factor $3/4$ ).

A B C X B1 B2 B3 B4 C' A'

Solution : Given a triangle ABC, we are required to construct another triangle whose sides are $3/4$ of the corresponding sides of the triangle ABC.

Steps of Construction :

1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

2. Locate $3/4$(the greater of 3 and 4 in 4 ) points $B_1$ , $B_2$ , $B_3$ and $B_4$ on BX so that $BB_1$ = $B_1B_2$ = $B_2B_3$ = $B_3B_4$.

3. Join $B_4C$ and draw a line through $B_3$ (the 3rd point, 3 being smaller of 3 and 4 in $3/4$) parallel to $B_4C$ to intersect BC at C′.

4. Draw a line through C′ parallel to the line CA to intersect BA at A′ (see Fig.).

Then, ∆A′BC′ is the required triangle.

Let us now see how this construction gives the required triangle.

By Construction 11.1, ${BC′}/{C′C}=3/1$

Therefore,

${BC}/{BC′}$=${BC′+C′C}/{BC′}$=$1+{C′C}/{BC′}$=$1+1/3$=$4/3$

Also C′A′ is parallel to CA. Therefore, ∆A′BC′ ~ ∆ABC.

So, ${A′B }/{AB }={A′C′ }/{AC} ={BC′ }/{BC}=3/4$

Example: Construct a triangle similar to a given triangle ABC with its sides equal to $5/3$ of the corresponding sides of the triangle ABC (i.e., of scale factor $5/3$ ) are $5/3$ of the corresponding sides of ∆ABC.

Steps of Construction :

1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

2. Locate 5 points (the greater of 5 and 3 in $5/3$) $B_1$, B2, $B_3$, $B_4$ and $B_5$ on BX so that $BB_1$ = $B_1B_2$ = $B_2B_3$ = $B_3B_4$ = $B_4B_5$.

3. Join $B3_$(the 3rd point, 3 being smaller of 3 and 5 in $5/3$ ) to C and draw a line through $B_5$ parallel to $B_3C$, intersecting the extended line segment BC at C′.

4. Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′ (see Fig. 11.4).

Then A′BC′ is the required triangle.

For justification of the construction, note that

∆ ABC ~ ∆ A′BC′.

Therefore, ${AB}/{A′B}$ =${AC}/{A′C′}$=${BC}/{BC′}$

But ${BC}/{BC′}$ =${BB_3}/{BB_5}$=$3/5$

So, ${BC′}/{BC}=5/3$ and therefore ${A′B }/{AB}$ = ${A′C′}/{AC}$ = ${BC′ }/{BC}$ = $5/3$