Question

Construct a triangle similar to the given $△ABC$, with its sides equal to $\frac{5}{3}$ of the corresponding sides of the triangle $ABC$.

Answer 1

Step I : Draw any ray BX making an acute angle with $BC$ on the side opposite to the vertex $A$.

Step II : From B cut off 5 ares
$B_1,B_2,B_3{B}_4$ and $B-5$ on $BX$ so that
${BB}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5$

Step III : Join $B_3$ to $C$ and dra a line through $B_5$ parallel to $B_{3}C$. intersecting the extended line segment $BC$ at $C$.

Step IV : Draw a line through $C$ parallel to $CA$ intersecting the extended line segment BA at A' Then A'BC' is the required triangle.

Justification :

Note that $△ABC\sim {\Delta }^{\prime }{BC}^{\prime }($ Since $AC\parallel A^{\prime }C)$

therefore,$\frac{AB}{AB}=\frac{AC}{AC}=\frac{BC}{BC}$

But, $\frac{BC}{BC}=\frac{{BB}_3}{{BB}_5}=\frac{3}{5}$,

Therefore $\cdot \frac{A^{\prime }B}{AB}=\frac{A^{\prime }{C}^{\prime }}{AC}=\frac{{BC}^{\prime }}{BC}=\frac{5}{3}$