Question

Let D and E be points on sides AB and AC of a triangle ABC such that
(A) DE is parallel to BC
(B) DE divides the area of the triangle ABC into two equal parts
The ratio of the distance from A to DE to the distance between DE and BC is

• A. 1:1
• B. 2:1
• C. $\sqrt{2}:1$
• D. $\sqrt{2}+1:1$

Answer 1

The correct option is D $\sqrt{2}+1:1$

Let $AP=x_1,PQ=x_2$
Now, $\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)}=\frac{1}{2}$
$\Rightarrow \frac{x_1^2}{(x_1+x_2{)}^2}=\frac{1}{2}$
$\Rightarrow 2x_1^2=x_1^2+x_2^2+2x_1{x}_2$
$\Rightarrow x_1^2-x_2^2-2x_1{x}_2=0$
$\Rightarrow \frac{x_1}{x_2}=\frac{2+\sqrt{4+4}}{2}$
$\Rightarrow \frac{x_1}{x_2}=\frac{\sqrt{2}+1}{1}$