Hans Walser, [20170501]

Parabola and
right triangle

Idea: Abdilkadir Altintas, Afyon, Turkey

We discuss a theorem connecting an arbitrary right triangle to a parabola.

Let *ABC* be a right triangle with right angle
at *C* (Fig. 1). *M* is the center of the circumcircle *c* and *F* the foot of the
perpendicular from *C* to the
hypotenuse *AB*. The line *d* is the tangent at the circumcircle *c* in the point *C*. Of course, the segment *MC*
is orthogonal to the tangent *c*.

Fig. 1: Right triangle

Fig. 2: Parabola

The
parabola *p* with focus *F* and directrix *d* passes through the points *A*
and *B* and is tangential to the legs *CA* and *CB* of the triangle *ABC*
(Fig. 2).

Fig. 3: Angles

The
triangle *CAM* is isosceles. Therefore
we have the angle .
Now we draw the perpendicular from *A*
to *d*. The perpendiculars *AF*_{1} and *MC* are parallel. Hence and .

Fig. 4: Congruent triangles

Now
the triangles *CAF* and *CAF*_{1} are congruent (same
angles, side *CA* in common) (Fig. 4).
The triangle *CAF*_{1} is just
the mirror image of the triangle *CAF*.

Therefore
the distance from *A* to *F* is equal to the distance from *A* to the line *d*. Hence *A* is a point of
the parabola *p*. The line *AC* is angle bisector of the angle and therefore
it is tangential to the parabola *p*.

Fig. 5: Argumentation on the other side

An
analogue argumentation holds for the point *B*
(Fig. 5).

Fig. 6: Tangent parallel to the hypotenuse

The
midpoint *D* of the segment *MC* is on the parabola *p*. The tangent in *D* at the parabola *p* is
parallel to the hypotenuse *AB*. The
proof is given by the isosceles triangle *FCD*.

Fig. 7: Intersection of three lines

Let *S* be the point of intersection of the
lines *d* and *AB,* and *E* the common
point of the parabola *p* and the line *FC*. The tangent *e* at *p* through *E* passes also through *S*. The proof derives from the isosceles
triangle *FGE*.