Proving the converse of the Pythagorean Theorem using slopes?

Question

I'm working through Calculus with Analytic Geometry by Simmons. I'm going through chapter 1 which is really just a pre-calculus review. I'm having issues with the following question:

Let $(0, 0)$, $(a, 0)$ and $(b, c)$ be the vertices of an arbitrary triangle placed so that one side lies along the positive $x$-axis with its left endpoint at the origin. If the square of this side equals the sum of the squares of the other two sides, use slopes to show that the triangle is a right triangle. Thus, the converse of the Pythagorean theorem is also true.

If $A = (0,0)$, $B = (a,0)$, and $C= (b,c)$, then we are given that ${d_{AB}}^2 = {d_{BC}}^2 + {d_{AC}}^2$.

I also know that for two lines to be perpendicular, the product of the slopes must equal $-1$ (i.e., the slopes of the two lines must be negative reciprocals of each other). I'm assuming using this fact is what Simmons means by using slopes.

I'm having a hard time seeing the relationship between ${d_{AB}}^2 = {d_{BC}}^2 + {d_{AC}}^2$ and the above fact about slopes and perpendicular lines. In other words, I'm not seeing how I am to use what is given to arrive at a solution using slopes. I am aware that the converse of the Pythagorean theorem is typically proved in a more geometrically pure way using proof by contradiction, but I was not able to find anything on the internet regarding the method that Simmons wants me to use.

Anything to help point me in the right direction would be much appreciated.

Answer 1

If the Pythatgorean theorem holds: $(b^2 + c^2) + ((a-b)^2 + c^2) = a^2$

As for the slopes, if they are at right angles $\frac cb\frac{c}{(b-a)} = -1$

Simplify and show that one implies the other.

Answer 2

The proof, for the sake of helping others if they ever need it:

Let $A = (0,0)$, $B = (a,0)$, and $C = (b,c)$. We are given that ${d_{AB}}^2 = {d_{BC}}^2 + {d_{AC}}^2$. Recall that for two lines to be perpendicular the slopes must be negative reciprocals of each other. Calculating the slopes of $\overleftrightarrow{BC}$ and $\overleftrightarrow{AC}$ we have:

$m_{BC} = \frac{c}{(b-a)}$ and $m_{AC} = \frac{c}{b}$.

So in order for $\triangle ABC$ to be a right triangle we must show that:

${d_{AB}}^2 = {d_{BC}}^2 + {d_{AC}}^2 \Rightarrow \frac{c}{(b-a)}\frac{c}{b} = -1$

Writing ${d_{AB}}^2 = {d_{BC}}^2 + {d_{AC}}^2$ in terms of $a$, $b$, and $c$ we have the following:

$a^2 = (c^2 + (a-b)^2) + (c^2 + b^2))$

Next, by simplifying:

$a^2 = (c^2 + (a-b)^2) + (c^2 + b^2)) \Rightarrow a^2 = 2c^2 + a^2 - 2ab + 2b^2 \Rightarrow -2b^2 = 2c^2 - 2ab$.

Solving for $a$ then gives:

$-2b^2 = 2c^2 - 2ab \Rightarrow -2(b^2 + c^2) = -2ab \Rightarrow b^2 + c^2 = ab \Rightarrow a = \frac{b^2 + c^2}{b}$.

Now substituting $a$ into $m_{BC}$ and simplifying yields the following:

$\frac{c}{(b-a)} = \frac{c}{(b - (\frac{b^2 + c^2}{b})} = \frac{c}{(-\frac{c^2}{b})} = \frac{-cb}{c^2} = \frac{-b}{c}$

Which we see is the negative reciprocal of $m_{AC}$:

$\frac{-b}{c}\frac{c}{b} = \frac{-bc}{bc} = -1.$

Therefore $\triangle ABC$ is a right triangle and the converse of the Pythagorean Theorem holds.