Areas in an isosceles triangle (corrected)

Let the area of $\triangle BEF$ be $a$ . Since $\triangle ABD$ is similar to $\triangle BEF$ and its sides are twice as long, the area of $\triangle ABD$ is $2^2 a = 4a$ . To generalize let the area of $\triangle BDC$ be $na$ , where $n$ is a positive integer, then the area of $\triangle ABC$ is $(n+4)a$ .

Then the ratio of the areas $\dfrac {[BDC]}{[ABC]} = \dfrac n{n+4}$ and we have:

$\begin{aligned} [BDC] & = \frac n{n+4} [ABC] \\ \frac 12 \overline{CD} \cdot \overline{BD} & = \frac n{n+4} \cdot \frac 12 \overline{BC} \cdot \overline{AC}\sin \theta \\ \overline{BC} \cos \theta \cdot \overline{BC} \sin \theta & = \frac n{n+4} \overline{BC} \cdot \overline{AC} \sin \theta \\ \overline{BC} \cos \theta & = \frac n{n+4} \overline{AC} \\ \frac {\overline{BC}}{\overline{AC}} \cos \theta & = \frac n{n+4} & \small \blue{\text{Note that }\frac {\overline{BC}}{\overline{AC}} = 2 \cos \theta} \\ 2 \cos^2 \theta & = \frac n{n+4} \\ \cos \theta & = \sqrt{\frac n{2(n+4)}} \\ \implies \theta & = \cos^{-1} \sqrt{\frac n{2(n+4)}} \end{aligned}$

For $n=10$ , we have $\theta = \cos^{-1} \sqrt{\dfrac 5{14}} \approx \boxed{53.3}^\circ$ .

Since $\triangle BFE \sim \triangle BAD$ by AA similarity, and since $BF = FA$ , $BE = ED$ .

Let $x = BE = ED$ . Then $CD = \frac{2x}{\tan \theta}$ , and the area of the blue triangle is $A_{\triangle BDC} = \frac{1}{2} \cdot CD \cdot BD = \frac{1}{2} \cdot \frac{2x}{\tan \theta} \cdot 2x = \frac{2x^2}{\tan \theta}$

From the angle sum of $\triangle CBD$ , $\angle CBD = 90° - \theta$ , and since $\angle ABC = \theta$ , $\angle FBE = \angle ABC - \angle CBD = \theta - (90° - \theta) = 2\theta - 90°$ . From the angle sum of $\triangle BFE$ , $\angle BFE = 90° - (2\theta - 90°) = 180° - 2\theta$ . That means $EF = \frac{x}{\tan (180° - 2\theta)} = \frac{-x}{\tan 2\theta}$ , and the area of the yellow triangle is $A_{\triangle BEF} = \frac{1}{2} \cdot BE \cdot EF = \frac{1}{2} \cdot x \cdot \frac{-x}{\tan 2\theta} = \frac{-x^2}{2\tan 2\theta}$ .

We are given that $A_{\triangle BDC} = 10 A_{\triangle BEF}$ , so $\frac{2x^2}{\tan \theta} = \frac{-10x^2}{2\tan 2\theta}$ . Substituting $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2 \theta}$ and solving gives $\theta = \tan^{-1} (\frac{3}{\sqrt{5}}) \approx \boxed{53.3°}$ .

As previously mentioned, the area of $\Delta ABD$ is $2^2 = 4$ times that of $\Delta BEF$ . Therefore, $\Delta BDC$ is only $\frac{10}{4} = 2.5$ times larger in area than $\Delta ABD$ .

Now, since $\Delta ABD$ and $\Delta DBC$ have the same base, $BD$ , their heights must be in the same ratio as their areas (this comes from the fact $A = \frac{1}{2}bh$ ). Let $AD = p$ . Then $CD = 2.5p$ . Since $ABC$ is isoceles, $AB = AC = p + 2.5p = 3.5p$ .

Using some trigonometry, $\tan \theta = \frac{BD}{2.5 p} \Rightarrow BD = 2.5 p \tan \theta$ . We also need to find $\angle BAC$ which is just $180º - 2 \theta$ . We then use the formula $A = \frac{1}{2}ab \sin C$ for a triangle to express $\Delta ABD$ and $\Delta BDC$ in terms of $\theta$ :

$\Delta ABD + \Delta BDC = \Delta ABC$ $\frac{1}{2} (AD)(BD) + \frac{1}{2} (BD)(DC) = \frac{1}{2} (AB)(AC) \sin(180º - 2 \theta)$ $\frac{1}{2} (p)( 2.5 p \tan \theta) + \frac{1}{2}(2.5 p \tan \theta)(2.5 p) = \frac{1}{2} (3.5p)(3.5p) \sin 2 \theta$ $2.5p^2 \tan \theta + 6.25 p^2 \tan \theta = 12.25 p^2 \sin 2 \theta$ $\tan \theta = 1.4 \sin 2 \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = 2.8 \sin \theta \cos \theta$ $1 = 2.8 \cos^2 \theta \Rightarrow \sqrt{\frac{1}{2.8}} = \cos \theta$ $\theta = \cos^{-1} \left( \sqrt{\frac{1}{2.8}} \right) = \boxed{53.3º}.$