# Three lines go through one point

The $ABDE$ and $BCFG$ squares are erected outwardly on sides $AB, BC$ of $\triangle ABC$. Prove that lines $FA, EC$ and the altitude of $\triangle ABC$ passing through $A$, intersect at one point. I tried forming a square and paralelogram, but it didn’t help. How should I prove it? Maybe with Ceva? I believe there exist a solution without using Ceva. Or van Aubel's theorem?

• What is "Ceva" -- some kind of proof assistant? Commented Jan 22, 2018 at 13:23
• en.wikipedia.org/wiki/Ceva%27s_theorem Commented Jan 22, 2018 at 13:29

Let us add a couple of additional points: $H$ and $I$, as the missing vertices of the square built on $AC$.

The solid green segments are orthogonal and with equal lengths due to a $90^\circ$ rotation around $C$. The solid purple segments are orthogonal and with equal lengths due to a $90^\circ$ rotation around $A$. Let us denote through $AF\cap CE$ as $Y$, and let $B'$ be the image of $B$ with respect to the translation bringing $H$ to $A$. $AF$ and $CE$ are heigths of $AB'C$, hence $Y$ lie on the height of $AB'C$ through $B'$. Since $B'BHA$ is a parallelogram by construction, $Y$ also lies on height of $ABC$ through $B$.

An alternative approach is to show through the cosine theorem that $YA^2-YC^2=BA^2-BC^2$ holds: the conclusion is the same, since the height through $B$ is exactly the locus of points $P$ such that $PA^2-PC^2=BA^2-BC^2$.

This classical result is related to Van Aubel's theorem for quadrilaterals.

Remark: if we denote $AG\cap CD$ as $X$, $BX$ goes through the center of $ACIH$, by a similar argument.

A complex number approach. We may assume without loss of generality that $C=0$, $A=1$ and $B=z=x+iy$. Hence $$F=C+i(B-C)=iz\quad,\quad E=A-i(B-A)=1-i(z-1),$$ and the intersection point $P$ of $CE$ and $FA$ is
$$P=sE+(1-s)C=tF+(1-t)A=x+i\frac{x(1-x)}{y+1}$$ with $s=x/(y+1)$ and $t=(1-x)/(y+1)$. Since $\text{Re}(P)=x$, it follows that $P$ is along the line through $B$ and orthogonal to $AC$, i.e. the altitude from $B$ to $AC$.

Let H be the foot of the perpendicular from B to AC. Let K be the foot of the perpendicular from E to the extension of AC - so KAC is a straight line, perpendicular to EK.

Look for similar triangles.

• How comes the fact that they pass through one point form similar triangles? Commented Jan 22, 2018 at 14:23
• Triangle EKC is cut by BH, and the part of BH inside the triangle is EKHC/KC=AHHC/(AC+BH), the same as the corresponding value when you swap A and C. So the two lines cut at the same place. Commented Jan 23, 2018 at 6:30