Why did my book write this over-complicated proof instead of the simpler one?

Question

Proof statement:

Three like parallel forces of magnitudes $P, Q$ and $R$ act at the vertices $A, B$ and $C$ of $\triangle ABC$, and $\frac{P}{a}=\frac{Q}{b}=\frac{R}{c}$, where $a, b$ and $c$ are the three sides of the triangle ($a$ is the side opposite to $\angle A$). Prove that their resultant passes through the incenter of the triangle.

My attempt:

Step 1:

Given,

$$\frac{P}{a}=\frac{Q}{b}=\frac{R}{c}$$

From the 2nd and the 3rd ratios,

$$\frac{Q}{b}=\frac{R}{c}$$

$$\frac{Q}{R}=\frac{b}{c}=\frac{AC}{AB}\tag{1}$$

Again,

$$\frac{Q}{CD}=\frac{R}{BD}=\frac{Q+R}{BC}$$

$$\text{Each force is proportional to the distance between the points of application of the other two.}$$

From the 1st and 2nd ratios,

$$\frac{Q}{CD}=\frac{R}{BD}$$

$$\frac{Q}{R}=\frac{CD}{BD}\tag{2}$$

From $(1)$ and $(2)$,

$$\frac{AC}{AB}=\frac{CD}{BD}$$

According to the angle bisector theorem, $AD$ is the bisector of $\angle A$.

Step 2:

Given,

$$\frac{P}{a}=\frac{Q}{b}=\frac{R}{c}$$

From the 1st and the 3rd ratios,

$$\frac{P}{a}=\frac{R}{c}$$

$$\frac{P}{R}=\frac{a}{c}=\frac{BC}{AB}\tag{3}$$

Again,

$$\frac{P}{CE}=\frac{R}{AE}=\frac{P+R}{AC}$$

$$\text{Each force is proportional to the distance between the points of application of the other two.}$$

From the 1st and 2nd ratios,

$$\frac{P}{CE}=\frac{R}{AE}$$

$$\frac{P}{R}=\frac{CE}{AE}\tag{4}$$

From $(3)$ and $(4)$,

$$\frac{BC}{AB}=\frac{CE}{AE}$$

According to the angle bisector theorem, $BE$ is the bisector of $\angle B$.

Step 3:

$AD$ and $BE$ intersect at $I$. So, $I$ is the incenter of $\triangle ABC$ (proved).

My book's attempt:

Step 1:

[Same as my step 1]

$AD$ is the bisector of $\angle A$.

Step 2:

Let

$$\frac{P}{a}=\frac{Q}{b}=\frac{R}{c}=k\tag{5}$$

Now,

Three forces $P$, $P+Q+R$, and $Q+R$ act on the line $AD$.

$$\frac{P}{DI}=\frac{Q+R}{AI}=\frac{P+Q+R}{AD}$$

$$\text{Each force is proportional to the distance between the points of application of the other two.}$$

From the 1st and 2nd ratios,

$$\frac{P}{DI}=\frac{Q+R}{AI}$$

$$\frac{AI}{DI}=\frac{Q+R}{P}$$

Now, using $(5)$,

$$\frac{AI}{DI}=\frac{bk+ck}{ak}=\frac{b+c}{a}\tag{6}$$

Again, from step 1,

$$\frac{AB}{BD}=\frac{AC}{CD}=\frac{AB+AC}{BD+CD}=\frac{b+c}{a}\tag{7}$$

From $(6)$ and $(7)$,

$$\frac{AB}{BD}=\frac{AI}{DI}$$

$BI$ is the bisector of $\angle B$.

Step 3:

$AD$ and $BI$ intersect at $I$. So, $I$ is the incenter of the triangle.

My comments:

I will agree that my book's proof is a valid proof. However, I feel it is overcomplicated and forced. On the other hand, my proof is simpler and feels more natural wherein step 2 is essentially a repeat of step 1. After completing the proof myself in my own way, when I saw my book's proof, I couldn't understand why my book didn't go for the simpler proof (my proof). Then I became worried as to whether my proof was correct or not. That's why I posted this question.

My question:

1. Is my attempt/proof correct?

Answer 1

Your procedure seems to be ok, hence I think nothing more relevant could be added.

Note the projection of each force into the closest sides of the triangle leads to a torque sum equal to zero from the incentre, since all the angles sum 180, the projections escalate properly due to the taken magnitudes $a$, $b$ and $c$, and the incentre being equidistant $r$ from each side. Could be a faster proof, but requires the Sum of Torque framework in place.

Note we are considering this Geometrical Corollary

Each force is proportional to the distance between the points of application of the other two.

You should remember all this comes Statics corollary, regarding the more general Principle of Statics, which is nothing more than Newton' Second Law. The last equation, is also known as Varignon Theorem, highlighting that the net momentum is equivalent to the momentum of the net force (which in Statics is zero). $$ F=\sum_{i=1}^n F_i=0 \\ T=\sum_{i=1}^n T_i=0 $$

As you could perhaps noted, as you are indirectly working under a Statics framework, the following holds

• The Corollary assumes a Segment is a Rigid Beam,
• The problem you solved assumes the connection between them are Rigid. In a Triangle, this is evident.
• Note how Statics also deals with Distributed Forces and Torques, in a way your book perhaps is not scoping. As distributions, a Punctual Load is a vector force, and a Distributed Load is... a function defining magnitudes of vector forces at each point. And back to the Punctual Load $F$, which are simply a Dirac Delta distribution $F\delta(p)$ at a Vector position $p$ (!).
• Not so far, when you assume your Beam is not Rigid, but just Stiff, some Tensors will appear instead of Forces. You will be curious in how about these Tensors work with all these previous objects, when they are allowed to "move".