2D geometry

Frame the problem

The ratio we hope to determine is given by the following\[ \frac{\mathrm{A}_{ABCD} - \mathrm{A}_{ADE}}{\mathrm{A}_{ABCD}} \] .If we call the angle ADE \(\theta\) then, according to the question, the angle ADC is \(2\theta\). We will refer to this angle, \(\theta\), in our workings below.

Area of the parallelogram ABCD

The area of the parallelogram ABCD is given by the product of base (DC) times perpendicular height, \(h_{1}\).Dropping a vertical line from vertex A we can see that the perpendicular height is given by\[ h_{1} = 3\sin\left( 2 \theta \right) \] .Therefore, the area of the parallelogram, \(\mathrm{A}_{ABCD}\) is given by:\[ \mathrm{A}_{ABCD} = 4 \times h_{1} = 12 \sin \left( 2 \theta \right) \] .By using the double angle formulae:\[ \sin \left( 2 \theta \right) = 2 \sin \left( \theta \right) \cos \left( \theta \right) \] ,we can express this area as follows:\[ \mathrm{A}_{ABCD} = 24 \sin \left( \theta \right) \cos \left( \theta \right) \] .NOTE: This double angle formula is just something that one should commit to memory.

Area of the triangle ADE

The area of the triangle ADE is also given by half the base (\(\frac{\mathrm{DE}}{2}\)) times the perpendicular height \(h_{2}\). In this case the perpendicular height runs from line DE to the vertex A.Given that the angle ADE is \(\theta\), as discussed above, then:\[ \frac{\mathrm{DE}}{2} = 3 \cos \left( \theta \right) \] ,and the perpendicular height, \(h_{2}\) will be \[ h_{2} = 3 \sin \left( \theta \right) \] . So the area of the triangle ADE is given as\[ \mathrm{A}_{ADE} = 3 \sin \left( \theta \right) \times 3 \cos \left( \theta \right) = 9 \sin \left( \theta \right) \cos \left( \theta \right) \] .

Combining for result

So plugging these expressions into our original ratio at the top\[ \frac{\mathrm{A}_{ABCD} - \mathrm{A}_{ADE}}{\mathrm{A}_{ABCD}} \]\[ = \frac{ 24 \sin \left( \theta \right) \cos \left( \theta \right) - 9 \sin \left( \theta \right) \cos \left( \theta \right) }{ 24 \sin \left( \theta \right) \cos \left( \theta \right) } \] .If we recognise \( \sin \left( \theta \right) \cos \left( \theta \right) \) as a common factor across numerator and denominator we can cancel that out to leave\[ \frac{ 24-9 }{24} = \frac{ 15 }{ 24} = \frac{ 5 }{8} \] .Hit me up on the comments if you see any errors!! Thanks.