Question

A line parallel to the base of a triangle cuts the triangle into two regions of equal area. This line also cuts the altitude into two parts. Find the ratio of the two parts of the altitude.

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

Answer 1

Answer: (D)

$1:(\sqrt{2}+1)$

Given that,

parallel line $DE$ divides $△ABC$ into two parts of equal area.

$\therefore Areaof△ADE=\frac{1}{2}\times Areaof△ABC$

$\frac{Areaof△ADE}{Areaof△ABC}=\frac{1}{2}$

$△ADE\sim △ABC$

We know that, the ratio of the area of two similar triangles is equal to the ratio of squares of the respective altitudes.

$\frac{Areaof△ADE}{Areaof△ABC}=\frac{A{M}^2}{A{L}^2}$

$\frac{Areaof△ADE}{Areaof△ABC}=\frac{A{M}^2}{(AM+ML{)}^2}$

$\frac{1}{2}=\frac{A{M}^2}{(AM+ML{)}^2}$

$\frac{2}{1}=\frac{(AM+ML{)}^2}{A{M}^2}$

$\sqrt{2}=\frac{(AM+ML)}{AM}$

$\sqrt{2}=1+\frac{ML}{AM}$

$\frac{ML}{AM}=\sqrt{2}-1$

$\frac{AM}{ML}=\frac{1}{\sqrt{2}-1}$

Multiply and divide RHS by $(\sqrt{2}+1)$

$\frac{AM}{ML}=\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$

$\frac{AM}{ML}=\frac{(\sqrt{2}+1)}{2-1}$

$\frac{AM}{ML}=\frac{(\sqrt{2}+1)}{1}$

$\frac{ML}{AM}=\frac{1}{(\sqrt{2}+1)}$

$ML:AM=1:\sqrt{2}+1$