# Very hard geometry problem.

Let $AB′C$ be a triangle with $\angle B′AC = 70^\circ$, $\angle ACB′ = 80^\circ$, $F$ is a point inside the triangle such that $\angle FB′A = \angle FCB` = 20^\circ$. Find $\angle AFB′$.

• Actually this does not deserve to be closed, as the OP did show some effort (writing extra numbers in angles as the OP calculated and let CF intersects AB at D), although that effort is made without words. May 28 '18 at 10:35
• Actually,@TrầnThúcMinhTrí, it does deserve to be closed since the OP didn't do a darn piece of work on it. This site is not intended for askers to ask problem statement questions: copied from an assigned question or a math-competition problem list, including copying the image, and fully expecting us to do their work for them, nor for users to answer them, @Oleg567. (It's a rather pathetic route to get easy rep by answering such questions.) May 28 '18 at 20:06
• @amWhy It seems to me that OP has done work on this question. The diagram included demonstrates work beyond the problem statement given: For instance the construction of points $D$ and $E$ which is an important step to the correct solution. For this reason, I'm voting to leave this question open. May 28 '18 at 23:42
• @B.Mehta: There is no work shown. Labeling a graph only is equivalent to typesetting a question to submit. Your argument could be applied to argue that "someone who posts the question: "Is $f: \mathbb R \to \mathbb R, \; f(x) = x^2\;$ a bijective function?" with a title "A very hard function problem", void all else, demonstrated sufficient effort because they typed the question to submit, and even used a little mathjax. There fore it should not be closed." That's ludicrous...(cont.) May 29 '18 at 13:50
• ...(cont.) Labeling a graph is no more work than identifying what $A, V$ are in the question "Show that $\frac{dV}{dt} = A(t), where V is the velocity, and A is acceleration." The final variable identifiers is not effort on the problem, itself. This question, like the example here, are expected clarifications, part of being clear about the bare problem statement imploring us to do the work for the asker. May 29 '18 at 13:56 ## 1 Answer Denote$h = EB'$,$a=EF$. Then $$EC = h\tan 10^{\circ},\quad AE = h\tan20^{\circ};$$ and other side: $$EC = a\tan 30^{\circ},\quad AE = a\tan x,$$ where$x=\angle AFE$; so $$h\tan 10^{\circ}=a\tan 30^{\circ}, \quad h\tan20^{\circ}=a\tan x;$$ therefore $$\tan x = \dfrac{h}{a}\tan 20^{\circ} = \dfrac{\tan 30^{\circ}}{\tan 10^{\circ}}\cdot \tan 20^{\circ}.$$ It remains to show that$x=50^{\circ}$and so$\angle AFB' = 180^{\circ}-x=130^{\circ}\$.

To see that $$\dfrac{\tan 20^{\circ} \tan 30^{\circ}}{\tan 10^{\circ}} = \tan 50^{\circ},$$ you can review this question+answers.

• But it can be proven by auxiliary constructions? May 28 '18 at 11:48
• @Pop Marius: I don't know; perhaps yes, but maybe in more complicated way. May 28 '18 at 12:00