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

Question

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:

Answer 1

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.

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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 $\angle APF\cong\angle ACE$. $\square$