# Prove that a triangle with given base and height, formed by the tangent of a parabola, shares an angle with a given right triangle.

Let $AB$ and $AC$ two perpendicular segments. On the line $\overleftrightarrow{AB}$ there is a point $D$ such that $\angle ACD =\alpha$. A line perpendicular to $\overleftrightarrow{CD}$ goes trhough A, and cuts it at the point $E$. Let $A'$ be the reflection of $A$ with respect of $E$. Then, trace a line trhough $A'$ that is parallel to $ED$.

Now, construct the parabola with focus $A$ and directrix $\overleftrightarrow{A'}$. trace a tangent line to this parabola from point $B$ and a line through $C$ that is parallel to $\overleftrightarrow{AB}$. Such lines intersect at point $F$.

Prove that $\angle AFB=\alpha$

Here's an image of the construction: Claim: $$\overleftrightarrow{CF}$$ is tangent to the parabola at, say, $$P$$. Moreover, $$\angle APF\cong \angle ACE$$. (Proof below.)

This actually allows a somewhat more profound version of the target result: Restatement. Let the tangents from a point $$F$$ meet a parabola with focus $$A$$ at points $$P$$ and $$P'$$. Then $$\angle APF\cong \angle AFP'$$ (and $$\angle AP'F\cong\angle AFP$$).

Reintroducing an element from the original statement, let the perpendicular from $$A$$ to $$\overleftrightarrow{FP}$$ meet that tangent at $$C$$; likewise, construct $$C'$$ on $$\overleftrightarrow{FP'}$$. (Note that $$C$$ and $$C'$$ lie on the parabola's vertex tangent.) By the Claim, $$\angle P\cong \angle C$$ and $$\angle P' \cong \angle C'$$. Now, for having opposing right angles, $$\square ACFC'$$ is cyclic. By the Inscribed Angle Theorem, $$\angle C \cong AFC'$$ (and $$\angle C'\cong \angle AFC$$) which gives the result. $$\square$$

Corollary. $$\overleftrightarrow{AF}$$ bisects $$\angle PAP'$$.

I wasn't aware of this properties described in the Restatement or the Corollary. I'll need to remember them.

Proof of Claim. Reflect $$A$$ about $$C$$ to get $$A''$$, which is necessarily on the directrix. Let the perpendicular raised from the directrix at $$A''$$ meet $$\overleftrightarrow{CF}$$ at $$P$$. Since $$\overleftrightarrow{CF}$$ is the perpendicular bisector of $$\overline{AA''}$$, we have $$\overline{AP}\cong\overline{A''P}$$, so that $$P$$ is, by definition, on the parabola. Moreover, $$\overleftrightarrow{CF}$$, as the bisector of $$\angle APA''$$, is tangent to that parabola. A little angle chasing (based on parallel segments $$\overline{AA'}$$ and $$\overline{PA''}$$, and isosceles triangle $$\triangle APA''$$) shows $$\angle APF\cong\angle ACE$$. $$\square$$