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diagramdiagram2

There is a right triangle $\textrm{ABC}$ like the diagram above, and the point $\textrm{D}$ set so that $\mathrm{\overline{AD}=\overline{BC}}$. If point $\mathrm{E}$ divides line segment $\mathrm{AB}$ in the ratio of $5:2$, $\mathrm{\overline{AD}=\overline{DE}=\overline{CE}}$. Let $\angle\mathrm{ABC}=\theta$, then find the value of $\tan\theta$.

So what I did was I selected a point $\mathrm{F}$ on $\mathrm{\overline{BC}}$ so that $\mathrm{\overline{EF}\perp\overline{BC}}$, then I let $\mathrm{\overline{EF}}=h$, $\mathrm{\overline{AD}}=x$.

Then I can say $$h:h+x=2:5$$ So I get $$h=\frac{2}{3}x$$

Then I can use pythagorean theorem on $\triangle\mathrm{EFC}$

I get $$\mathrm{\overline{FC}}=\frac{\sqrt{5}}{3}x$$

Therefore $$\tan\theta=\frac{\mathrm{\overline{FE}}}{\mathrm{\overline{BF}}}=\frac{2}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{2}$$

However, If I just find $$\tan\theta=\frac{\mathrm{\overline{CA}}}{\mathrm{\overline{BC}}}=\frac{7}{3}$$

I am really confused. Is there something wrong with my steps?

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    $\begingroup$ @MatthewDaly I meant $h:2h+x=2:7$ but I just did $h:h+x=2:5$ considering that $\endgroup$ – Pizzaroot Aug 17 at 18:26
  • $\begingroup$ Hmmmm, I'm setting this up in Geogebra and am getting different results than both of yours and Aqua's as well. This problem is cursed! $\endgroup$ – Matthew Daly Aug 17 at 18:39
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So $<BEC = \theta$ and thus $<BCE = 180^{\circ}-2\theta$

So $<ECD= 2\theta-90^{\circ} = <EDC$ and thus $<CED = 360^{\circ}-4\theta $

So $<DEA = 3\theta -180^{\circ} = < EAD$

So we have: $$\theta +(3\theta -180 ^{\circ}) = 90^{\circ} \implies \theta = 67,5^{\circ}$$

So $$\tan \theta = 1+\sqrt{2}$$

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    $\begingroup$ But still doesn't solve my problem; So does this mean that the problem has too many information that doesn't fit together? $\endgroup$ – Pizzaroot Aug 17 at 18:42
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    $\begingroup$ Yes, information $5:2$ does not hold, it should be refuted. $\endgroup$ – Aqua Aug 17 at 18:44
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    $\begingroup$ yeah, 5:2 and right angle don't add up $\endgroup$ – Matthew Daly Aug 17 at 18:45
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There is no such triangle.

By angle-chasing, $\theta=\frac{3}{8}\pi$, which yields $\tan(\theta)=1+\sqrt{2}$.

But then, since $\theta$ is uniquely determined, the diagram is already forced, up to similarity, so the ratio $AE{\,:\,}\!BE$ is uniquely determined, and can't just be arbitrarily specified as $5{\,:\,}2$ (unless it actually is $5{\,:\,}2$).

Given the known value of $\theta$, let's find the ratio $AE{\,:\,}\!BE$ . . .

Without loss of generality, we can assume $BC=1$.

Then by right-triangle trigonometry, we get $$ AB=\text{sec}(\theta) $$ and by the law of sines in triangle $BCE$, we get $$ \frac{BE}{\sin(\pi-2\theta)}=\frac{1}{\sin(\theta)}$$ which yields $$BE=\frac{\sin(\pi-2\theta)}{\sin(\theta)}=\frac{\sin(2\theta)}{\sin(\theta)}=2\cos(\theta)$$ hence \begin{align*} \frac{AE}{BE}&=\frac{AB-BE}{BE}\\[4pt] &=\frac{AB}{BE}-1\\[4pt] &=\frac{\text{sec}(\theta)}{2\cos(\theta)}-1\\[4pt] &=\frac{\text{sec}^2(\theta)-2}{2}\\[4pt] &=\frac{\tan^2(\theta)-1}{2}\\[4pt] &=\frac{\left(1+\sqrt{2}\right)^2-1}{2}\\[4pt] &=1+\sqrt{2}\\[4pt] \end{align*} hence $AE{\,:\,}\!BE\ne 5{\,:\,}2$.

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