# Ravi’s substitution for sides of a triangle

Theorem : Let $a,b,c$be three real numbers. Then they are the sides of a triangle if and only if there are positive real numbers $x,y,z$such that $\boxed{\boxed{a=x+y}, \boxed{b=y+z}, \boxed{c=z+x}}$.

For the first direction : If $x,y,z$ are positive real numbers then clearly $x,+y, y+z, z+x$ form side lengths of a triangle.

For the other direction : If $a,b,c$ form side lengths of a triangle then we put $\boxed{x = \frac{a+c-b}{2}}, \boxed{y= \frac{a+b-c}{2}}, \boxed{z= \frac{b+c-a}{2}}$. Then easily we see that, $x>0,y>0,z>0$ and $\boxed{ \boxed{a=x+y}, \boxed{b=y+z}, \boxed{c=z+x}}$.

This completes the proof of the theorem $\blacksquare$.

Another way of proving (this is more natural and beautiful) the latter part is as follows: Let $\triangle ABC$ have side lengths $a,b,c$. Now consider the incircle (a.k.a. the inscribed circle) of  $\triangle ABC$. Let $P,Q,R$ be the points on the sides of $\triangle ABC$ where the incircle touches the sides of the triangle. Then using the fact that tangents drawn from an external point to a circle are equal in length gives the desired result.

This simple theorem can be used to solve many problems (especially from mathematical inequalities and geometry).

For example you can try to use it to solve the following problem :

If $a,b,c$ are side lengths of a triangle then show that $\frac{1}{a+b}, \frac{1}{b+c}, \frac{1}{c+a}$ must also be side lengths of a triangle.

# Willy's Lemma

Lemma : If $a,b,x,y$ be four acute angles such that $a+b=x+y$ and $\frac{ \sin (a) }{ \sin (b) } = \frac{ \sin (x) }{ \sin (y) }$ then $a=x$ and $b=y$.

It can be proved using simple trigonometric argument using the equation that : $\boxed{ 2 \sin (k) \sin (t) = \cos (k-t) - \cos (k+t) }$, and then desired result follows from the injectivity of the cosine function on $(0, \pi )$

There is another synthetic proof using simple geometry which I find quite interesting (I came to know about it from my friend Carlos). I present it here.

Consider an isosceles triangle $ABC$ such that $\angle BAC=a+b=x+y$ and $AB=AC$

Consider a point $D$ on $BC$ such that $\angle BAD= \angle a$ and $\angle CAD= \angle b$

Consider a point $E$ on $BC$ such that $\angle BAE= \angle x$ and $\angle CAE= \angle y$

Apply the general angle bisector theorem, and by the given conditions, we get:

$\frac{AD}{BD} = \frac{AE}{CE}$

$\Longrightarrow E \equiv D$
$\Longrightarrow \angle a= \angle x$ and $\angle b= \angle y$

This completes the proof.

#geometry