Question

Draw a triangle $\mathrm{ABC}$ with side $\mathrm{BC}=7\mathrm{cm},\angle \mathrm{B}=45°,\angle \mathrm{A}=105°$. Then, construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $∆ABC$. Give the justification of the construction.

Answer 1

Step 1. Draw a triangle $\mathrm{ABC}$ with side $\mathrm{BC}=7\mathrm{cm},\angle \mathrm{B}=45°,\angle \mathrm{A}=105°$.

The Sum of all interior angles of a triangle is $180°$.

$\begin{array}{rcl}\angle A+\angle \mathrm{B}+\angle \mathrm{C}& =& 180°\\ 105°+45°+\angle \mathrm{C}& =& 180°\\ \angle \mathrm{C}& =& 180°-\left(105°+45°\right)\\ \angle \mathrm{C}& =& 30°\end{array}$

Draw a base $\mathrm{BC}$ of the side $7\mathrm{cm}$:

Draw $\angle \mathrm{B}=45°$ and $\angle C=30°$:

Let the point $A$ be the point where the two rays intersect:

$△\mathrm{ABC}$ is formed.

Step 2. Draw a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $∆ABC$.

Draw a ray $\mathrm{BX}$ making an acute angle with side $\mathrm{BC}$ on the opposite side of a vertex $\mathrm{A}$:

Locate $4$ points ${\mathrm{B}}_1,{\mathrm{B}}_2,{\mathrm{B}}_3$ and ${\mathrm{B}}_4$ on $\mathrm{BX}$ where ${\mathrm{BB}}_1={\mathrm{B}}_1{\mathrm{B}}_2={\mathrm{B}}_2{\mathrm{B}}_3={\mathrm{B}}_3{\mathrm{B}}_4$:

Join ${\mathrm{B}}_3\mathrm{C}$:

Draw a line through ${\mathrm{B}}_4$ parallel to ${\mathrm{B}}_3\mathrm{C}$ intersecting extending $\mathrm{BC}$ to $\mathrm{C}\text{'}$:

Draw a line from parallel to $\mathrm{AC}$ intersecting extending $\mathrm{AB}$ to $\mathrm{A}\text{'}$:

$△A\text{'}BC\text{'}$ is formed.

Step 3. Justification of the construction.

It is justified when we prove $\mathrm{A}\text{'}\mathrm{B}=\frac{4}{3}\mathrm{AB},\mathrm{BC}\text{'}=\frac{4}{3}\mathrm{BC},$ and $\mathrm{A}\text{'}\mathrm{C}\text{'}=\frac{4}{3}\mathrm{AC}$.

In $△\mathrm{ABC}$ and $△\mathrm{A}\text{'}\mathrm{BC}\text{'}$.

$$\begin{gathered} \begin{array}{rcl}\angle \mathrm{ABC}& =& \angle \mathrm{A}\text{'}\mathrm{BC}\text{'}\left[\mathrm{Common}\right]\end{array} \\ \begin{array}{rcl}\angle \mathrm{ACB}& =& \angle \mathrm{A}\text{'}\mathrm{C}\text{'}\mathrm{B}\left[\mathrm{Corresponding}\mathrm{angle}\right]\end{array} \\ \because △\mathrm{ABC}~△\mathrm{A}\text{'}\mathrm{BC}\text{'}\left[\mathrm{AA}\mathrm{similarity}\mathrm{criterion}\right] \\ \therefore △\mathrm{ABC}~△\mathrm{A}\text{'}\mathrm{BC}\text{'} \\ \begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{A}\text{'}\mathrm{B}}& =& \frac{\mathrm{BC}}{\mathrm{BC}\text{'}}=\frac{\mathrm{AC}}{\mathrm{A}\text{'}\mathrm{C}\text{'}}\dots \left(1\right)\end{array} \end{gathered}$$

In $△{\mathrm{BB}}_3\mathrm{C}$ and $△{\mathrm{BB}}_4\mathrm{C}\text{'}$.

$\begin{array}{rcl}\angle {\mathrm{B}}_3\mathrm{BC}& =& \angle {\mathrm{B}}_4\mathrm{BC}\text{'}\left[\mathrm{Common}\right]\\ \angle {\mathrm{BB}}_3\mathrm{C}& =& \angle {\mathrm{BB}}_4\mathrm{C}\text{'}\left[\mathrm{Corresponding}\mathrm{angles}\right]\\ & & \therefore △{\mathrm{BB}}_3\mathrm{C}~△{\mathrm{BB}}_4\mathrm{C}\text{'}\left[\mathrm{AA}\mathrm{similarity}\mathrm{criterion}\right]\\ & & \because △{\mathrm{BB}}_3\mathrm{C}~△{\mathrm{BB}}_4\mathrm{C}\text{'}\\ \frac{\mathrm{BC}}{\mathrm{BC}\text{'}}& =& \frac{{\mathrm{BB}}_3}{{\mathrm{BB}}_4}\\ \frac{\mathrm{BC}}{\mathrm{BC}\text{'}}& =& \frac{3}{4}\dots \left(2\right)\end{array}$

From $\left(1\right)$ and $\left(2\right)$, we get

$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{A}\text{'}\mathrm{B}}& =& \frac{\mathrm{BC}}{\mathrm{BC}\text{'}}=\frac{\mathrm{AC}}{\mathrm{A}\text{'}\mathrm{C}\text{'}}=\frac{3}{4}\end{array}$

$\mathrm{A}\text{'}\mathrm{B}=\frac{4}{3}\mathrm{AB},\mathrm{BC}\text{'}=\frac{4}{3}\mathrm{BC},$ and $\mathrm{A}\text{'}\mathrm{C}\text{'}=\frac{4}{3}\mathrm{AC}$.

Hence, the sides of the constructed triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $∆ABC$.