Question

A line parallel to the base of a triangle cuts the 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.

Answer 1

Given that
parallel line DE divides ΔABC into two parts of equal area.

Therefore, 
Area of ΔADE = 1/2 * Area of ΔABC

$\frac{\Delta ADE}{\Delta ABC} = \frac{1}{2}$

ΔADE ~ ΔABC

We know that, the ratio of the area of two similar triangles is equal to the ratio of squares of the respective altitudes.
Therefore, 
$\frac{\Delta ADE}{\Delta ABC} = \frac{AM^2}{AL^2}$

$\frac{\Delta ADE}{\Delta ABC} = \frac{AM^2}{(AM+ML)^2}$

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

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

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

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

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

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

AM : ML = 1 : $\sqrt{2} -1}$

Answer 2

See diagram.

The area of triangle Δ AEF = Quadrilateral EBCF.

$\frac{1}{2} EF * AD\ = \frac{1}{2} (EF + BC) DG \\ \\ EF * AD = (EF + BC) DG \\ \\ \frac {AD}{DG} = \frac {EF + BC}{EF} = 1 + \frac{BC}{EF}\ \ -- Equation 1 \\ \\ \\ As\ ABC\ and\ AEF\ are\ similar,\ \\ \\ \frac{BC}{EF}\ =\ \frac{AG}{AD}\ =\ \frac{(AD+DG)}{AD}\ =\ 1\ +\ \frac{DG}{AD}\ \ --\ Equation 2 \\ \\ Call\ \frac{AD}{DG}\ =\ x\ =\ Ratio\ of\ two\ parts\ of\ the\ Altitude.$

$\\ \\ From\ equations\ 1\ and\ 2\ we\ get \\ \\ x = 1 + \frac{BC}{EF} = 1 + 1 + \frac{1}{x} = 2 + \frac{1}{x} \\ \\ x^2 - 2\ x - 1 = 0 \\ \\ x = 1 +- \sqrt2 => x = (\sqrt2 - 1),\ as\ x\ is\ positive. \\ \\ \ Ratio: \sqrt{2} - 1$