Question

Consider an acute angled ΔABC. D is any point on the side BC such that $\frac{ar\left(△ABD\right)}{ar\left(△ADC\right)}$ = $\frac{2}{3}$. How would you proceed to construct this line segment AD?

• A. Draw the median of the $△ABC$
• B. Draw a ray BX which makes an acute angle with side BC.
• C. Construct triangle ADC such that $△ADC\sim △ABC$
• D. Construct triangle ABD such that $△ABD\sim △ABC$

Answer 1

The correct option is C

Construct triangle ADC such that $△ADC\sim △ABC$

$\frac{ar\left(△ABD\right)}{ar\left(△ADC\right)}$ = $\frac{\left(\frac{1}{2}\right)\left(baseBD\right)\left(heightAE\right)}{\left(\frac{1}{2}\right)\left(baseDC\right)\left(heightAE\right)}$ = $\frac{BD}{DC}$ = $\frac{2}{3}$

So, D divides BC in the ratio 2:3. Our next obvious task, is therefore to find this point D. So, we will draw a ray BX which makes an acute angle with side BC. Now, perform the constructions required to mark D on BC such that BD:DC = 2:3.

The steps of construction:

First you need to construct the triangle ABC. Now that you have constructed the triangle ABC, we can divide the Line BC such that $\frac{ar\left(△ABD\right)}{ar\left(△ADC\right)}$ = $\frac{2}{3}$. From the solution, we found out that $\frac{BD}{DC}$ = $\frac{2}{3}$.
a) Let us start by constructing a ray BX which makes an acute angle with line segment BC
b) Now mark $B_1$,$B_2$,$B_3$,$B_4$ and $B_5$ on ray BX such that $B_1$$B_2$ = $B_2$$B_3$ = $B_3$$B_4$ = $B_4$$B_5$. (This is done in order to divide $B{B}_2$ and $B_2$$B_5$ in the ratio 2:3)
c) Now join $B_{5}C$ and draw $B_{2}D$ parallel to $B_{5}C$ such that the point D lies on the line segment BC. By Basic Proportionality Theorem, $\frac{BD}{DC}$ = $\frac{B{B}_2}{B_2{B}_5}$ = $\frac{2}{5}$.
d) Join the points A and D.

The ratio of the area of the resulting triangles $△$ABD and $△$ADC will be $\frac{2}{3}$.