Point of intersections of perpendiculars from a vertex of a triangle to the angle bisectors are collinear

Question

Prove that the feet of the four perpendiculars dropped from a vertex of a triangle upon the four angle bisectors of the other two angles are collinear.

Drawing a diagram, it is clear that $AFBE$ and $AGCD$ are rectangles. Also $DE \parallel BC \parallel FG$. However I can't prove that $DE$ and $FG$ are the same line. Any hints?

Answer 1

Let the angle bisectors at $A$ meet the perpendiculars from $C$ at $P$ and $Q$. Then $\square APCQ$ is a rectangle, which we can adorn as follows:

Since $\angle PAB \cong \angle APQ$, we have $\overline{PQ}\parallel\overline{AB}$. Moreover, since $\overline{PQ}$ meets $\overline{AC}$ at their respective midpoints, it is collinear with the corresponding midsegment of $\triangle ABC$.

The same is true for the corresponding figure (not shown) relative to vertex $B$, and the result follows. $\square$

Answer 2

A way of solving: Make side $BC$ be in the x-axis so you have the coordinates $B(0,0),C(a,0),A(a_1,a_2)$. A proof is given by

(1) Calculate the four points $F,D,E,G$.

(2) Write the equation of the $L$ line passing through two points of (1).

3) Verify that the other two points belong to $L$ line. $$***$$

1)► Angle bisectors of $\angle {ABC}$.

Put $d=\sqrt{a_1^2+a_2^2}$ so you have the two ones $$\frac{|a_1y-a_2x|}{d}=|y|$$

BE: $y=\dfrac{a_2x}{a_1-d}$ and BF: $y=\dfrac{a_2x}{a_1+d}$

► Angle bisectors of $\angle{BCA}$

Put $d_1=\sqrt{a_2^2+(a-a_1)^2}$ so you have $$\frac{|a_2x+(a-a_1)y-aa_1|}{d_1}=|y|$$ CD: $y=\dfrac{-a_2x}{a-a_1-d_1}+\dfrac{aa_1}{a-a_1-d_1}$ and CG: $y=-\dfrac{-a_2x}{a-a_1+d_1}+\dfrac{aa_1}{a-a_1+d_1}$

► Lines AE, AF, AD and AG.

AE: $\dfrac{y-a_2}{x-a_1}=\text{ the same pente as BF }=\dfrac{a_2}{a_1+d}$

AF: $\dfrac{y-a_2}{x-a_1}=\text{ the same pente as BE }=\dfrac{a_2}{a_1-d}$

AD: $\dfrac{y-a_2}{x-a_1}=\text{ the same pente as CG }=?$

AG: $\dfrac{y-a_2}{x-a_1}=\text{ the same pente as CD }=?$

NOTE.- Obviously you could find these pentes considering perpendicularity to known lines but why not using the already written pentes elsewhere?

You have clearly $F=BF\cap AF,\space D=CD\cap AD,\space E=BE\cap AE,\space G=CG\cap AG$ solved by easy linear systems.

Now follow (2) and (3).