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 ${A′B }/{AB}$ = ${A′C′}/{AC}$ = ${BC′ }/{BC}$ = $5/3$