# Ratio of areas of two triangles

Based on the figure below, what is the ratio of the area of triangle $CGI$ to the area of triangle $ABC$, in terms of $\theta$?

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All the right triangles in the picture are similar, so (noting $|BG|=|BD|$)

\begin{align} \frac{|\triangle CIG|}{|\triangle ABC|} &= \left(\frac{|CG|}{|AC|}\right)^2 = \left(\frac{|BC|-|BG|}{|AC|}\right)^2 = \left(\frac{|BC|-|BD|}{|AC|}\right)^2 \\ &= \left(\frac{|BC|\left(1-\sin\theta\right)}{|AC|}\right)^2 = \left(\frac{|BC|}{|AC|}\right)^2\left(1-\sin\theta\right)^2 \\ &= \tan^2\theta \left(1-\sin\theta\right)^2 \end{align}

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Fixing $AC$ at length $1$, $\dfrac{\text{Area}(CGI)}{\text{Area}(ABC)}$ is seen to be $\dfrac{\left ( {\dfrac{\sin(\theta)}{\cos(\theta)}} - \sin(\theta)\tan(\theta)\right )^2 \tan(\theta)}{\sin(\theta) \left (\cos(\theta)+ \sin(\theta)\tan(\theta) \right)}$. Mathematica simplifies this ratio to $(\sin(\theta)-1)^2\tan^2(\theta)$.

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To whomever down-voted this answer: can you explain why you down-voted? –  idonutunderstand Jan 27 '13 at 7:36
Looks okay to me. –  Brian M. Scott Jan 27 '13 at 7:44

Mathematica isn’t needed for the simplification, which is quite straightforward:

\begin{align*} \frac{\left({\sin\theta \over \cos\theta} - \sin\theta\tan\theta\right)^2 \tan\theta}{\sin\theta\big(\cos\theta+\sin\theta\tan\theta\big)}&=\frac{\left(\tan\theta-\sin\theta\tan\theta\right)^2\tan\theta}{\frac{\sin\theta}{\cos\theta}\left(\cos^2\theta+\sin^2\theta\right)}\\\\ &=\frac{\tan^3\theta(1-\sin\theta)^2}{\tan\theta}\\\\ &=\tan^2\theta(1-\sin\theta)^2\;. \end{align*}

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