Proving Three Lines in a Triangle Meet – Methods and Proofs

Question

A right angle triangle $ABC$ with squares formed by the side lengths $a, b, c$ is given.

The lines $AK$ and $BF$ are drawn as per the image crossing sides $a$ and $b$ respectivley in their respective points $A'$ and $B'$.

A third line is drawn from $C$ through side $c$ crossing it in point $M$ such that angle $\angle AMC$ is equal to $90$ degrees.

Prove that these lines intersect each other for any right angle triangle.

I belive the intended solution uses Ceva's theorem.

If we apply it to this given shape we must prove that

$$
(BA'/CA')\cdot (CB'/AB')\cdot (AM/BM)=1\ ,
$$
but i do not where to go from here.

I could through expressing the lines $AK$, $BF$ and $CM$ as functions establish the positions of $A'$, $B'$ and $M$ in sides $a$ and $b$ but could not get further in the given time for my test.

I do see that there is a possibility of solving the question with equal shapes as $ABC$, $BCM$ and $ACM$ share the same angles.

Note that no measurments are given and that it is to be solved for any right angle triangle and that no amount of trigonometry is allowed (not anything that would require a calculator), only geometry.

Any help is appreciated.

Geogebra representation of the question at hand:

No measurements were given.

Answer 1

In the notations of the posted picture we have, also considering signs: $$ \begin{aligned} -\frac{A'B}{A'C} &= \frac{BA'}{A'C} = \frac{BK}{AC}= \frac{BC}{AC} &(\Delta A'BK &\sim\Delta A'CA)\ ,\\[5mm] -\frac{B'C}{B'A} &= \frac{CB'}{B'A} = \frac{BC}{AF} = \frac{BC}{AC} &(\Delta B'CB &\sim\Delta B'AF)\ ,\\[5mm] -\frac{MA}{MB} &= \frac{AM}{MB} = \frac{AM\cdot AB}{MB\cdot AB} = \frac{AC^2}{BC^2} \ . \\[3mm] &\qquad\text{ All together:} \\[3mm] \frac{A'B}{A'C}\cdot \frac{B'C}{B'A}\cdot \frac{MA}{MB} &=-\frac{BC}{AC}\cdot \frac{BC}{AC}\cdot \frac{AC^2}{BC^2} =-1\ . \end{aligned} $$ Using Ceva's theorem, the lines $AA'$, $BB'$, $CM$ are concurrent.

$\square$

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We are done with the given problem in the given very special situation, but for the sake of the geometric taste, let us place the result in a more general context.

Generalization:

Proposition: Let $ABC$ be a triangle, on the sides are constructed in the exterior of the triangle squares. Among the many vertices we need only two, $F,K$, the new vertices of the two squares on $CA$, respectively $CB$, so that $FA\perp CA$ and $KB\perp CB$. Let $M$ be the projection of $C$ on $AB$, so that $CM$ is a height of $\Delta ABC$.

Then the lines $AK$, $BF$, $CM$ are concurrent.

Before we start the proof, recall the following theorem of Pappus. It deals with two lines, on the one line we have the points, in order, $A,B,C$, on the other line corresponding points $a,b,c$, also in the written order. Now we build intersections $X,Y,Z$ (see the linked wiki page), and the theorem insures that they are collinear.

In projective geometry it is also allowed, to use points at infinity. Such a point is given by specifying a (classical) line $l$, and then its point at infinity, denoted here by $\infty_l$, is "the limit point" (in a suitable model) when we move a point on $l$ in some chosen coordinate system on the line to infinity. Incidence properties are important only in projective geometry. An other line $k$ also goes through $\infty_l$ iff $k$ and $l$ are parallel / have same direction.

Proof of the Proposition:

With the above in mind, we can apply Pappus for our configuration, below in picture a general case:

Here, $f$ is the line $AF$, $k$ is the line $AK$, and we apply the theorem for:

• the points $A,F,\infty_f$ on the line $f$,
• the points $B,K,\infty_k$ on the line $k$. Doing so we build the three intersection points:
• $BF\cap AK$, this is the green point.
• $B\infty_f\cap A\infty_k=H$, which is the orthocenter of the triangle $\Delta ABC$, since $B\infty_f\|f$ is perpendicular on the side $AC$, and $A\infty_k\| k$ is perpendicular on the side $BC$.
• $K\infty_f\cap A\infty_k=H$,

and it remains to show that $TH\perp AB$, because then this line passing through the orthocenter $H$ is the same as $CM$. For this, let us denote by $a,b,c$ the sides of $\Delta ABC$, by $A,B,C$ also abusively the angles in the same vertices, and by $h_a,h_b,h_c$ the heights (each seen as line, and as length, depending on the context).

We compute the slope of $TH$, in the system with axes parallel to the lines $AC$ and $BC$. The projection of $KB$ on $AC$ is after a $90^\circ$-rotation the same as the one of $BC$ (same length as $AC$) on the perpendicular direction, so it is $h_b$. Its projection on $BC$ is $h_a$.

The segment $CH$ has proportional projections, $CH\cdot \cos\widehat {HCA}=CH\sin A$ on $AC$, and $CH\sin B$ on $BC$. We check this proportionality in one line, using: $$ \frac{CH\sin A}{CH\sin B} = \frac{\sin A}{\sin B} = \frac ab = \frac{h_b}{h_a} \ . $$ (Last equality uses the formula for the area, so $ah_a/2=bh_b/2$.)

This shows that $CH$ and $TH$ have same direction, the direction perpendicular to $AB$. This concludes the proof of the Proposition.

$\square$

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Note: Applying Pappus in the same manner on the initial configuration is much easier, almost a two lines proof.