How to calculate the projection of a side in a triangle

Question

Given the following triangle, with the vertices $P, +q, -q$

why is $r_2 - r_1$ the projection of the side $-q+q$ on the side $-qP$?

Thanks a lot !

Answer 1

When $\theta=0$, it is evident that the projection is equal to $r_2-r_1$. This is the only case in which this relation is verified.

In fact, this projection, noted $X$, is equal to $a\space cos\space x$ where $x$ is the angle at the vertex $-q$; besides $cos\space x=\frac{a^2+r_2^2-r_1^2}{2ar_2}$.

Therefore $$X= \frac {a(a^2+r_2^2-r_1^2)}{2ar_2}=r_2-r_1\Rightarrow a^2=(r_2-r_1)^2$$ This means $a=r_2-r_1$. But this is impossible because the resulting isosceles triangle would then have two right angles.Or more clearly, because we would have a right triangle in which the hypotenuse would be equal to a leg.