# Why am I getting two different answer for this geometry question?

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?

• @MatthewDaly I meant $h:2h+x=2:7$ but I just did $h:h+x=2:5$ considering that – Pizzaroot Aug 17 at 18:26
• 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! – Matthew Daly Aug 17 at 18:39

So $$ and thus $$

So $$ and thus $$

So $$

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

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

• But still doesn't solve my problem; So does this mean that the problem has too many information that doesn't fit together? – Pizzaroot Aug 17 at 18:42
• Yes, information $5:2$ does not hold, it should be refuted. – Aqua Aug 17 at 18:44
• yeah, 5:2 and right angle don't add up – Matthew Daly Aug 17 at 18:45

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 $$AE{\,:\,}\!BE\ne 5{\,:\,}2$$.