Prove that $\frac{ab}{a+b+c}= \frac{a+b-c}{2}$ if $a$ and $b$ are the legs of a right triangle and $c$ is the hypotenuse.

Question

I am working on a problem to prove that the radius of the incircle of a right triangle is $\frac{a+b-c}{2}$. Using the formula radius * semiperimeter = area, I have solved to where the radius = $\frac{ab}{a+b+c}$. How to manipulate or use Power of Point to come to this answer?

Answer 1

Use the fact that\begin{align}\frac{ab}{a+b+c}=\frac{a+b-c}2&\iff2ab=(a+b+c)(a+b-c)\\&\iff2ab=(a+b)^2-c^2\\&\iff a^2+b^2=c^2.\end{align}

Answer 2

We wish to show that if $a$, $b$, and $c$ are the sides of a right triangle with $c$ the hypotenuse, then

$$\cfrac{ab}{a+b+c}=\cfrac{a+b-c}{2}.$$

Notice that this is equivalent to $$2ab=(a+b+c)(a+b-c)=(a+b)^{2}-c^{2},$$ by difference of squares. And this equation, via expansion of $(a+b)^{2}$, is equivalent to $$2ab=a^{2}+b^{2}+2ab-c^{2},$$ or $$a^{2}+b^{2}=c^{2},$$ which is of course true by the Pythagorean Theorem. All our steps are reversible so we're done.

Answer 3

For a purely geometric proof of the desired formula for $r$, the radius of the incircle:

Let $A,B,C$ be the three vertices of the triangle (With $A$ opposite the side of length $a$, etc.). Let $P_a,P_b,P_c$ be the three points of tangency between the incircle and the triangle (with $P_a$ on the side of length $a$, etc.)

It is easy to see that $\overline {CP_a}=\overline {CP_b}=r$

(N.B. this is where we use the fact that we have a right triangle, with the right angle at $C$).

It follows that $\overline {AP_b}=b-r$ and $\overline {BP_a}=a-r$.

It then follows that $\overline {AP_c}=b-r$ and $\overline {BP_c}=a-r$ since the two tangents from a point outside a circle to the circle have the same length. But then $$c=\overline {AP_c}+\overline {BP_c}=a+b-2r$$ and the desired result follows immediately.

Answer 4

If that's even true it must be compatible with the most basic and famous of results about right triangles that $a^2 + b^2 = c^2$.

So ... just muck about till you get it.

$\frac{ab}{a+b+c}= \frac{a+b-c}{2} \iff$

$\frac {2ab}{a+b+c} = a + b -c \iff$

$2ab = (a+b+c)(a+b-c) = a^2 + ab -ac + ab + b^2 -bc +ac +bc - c^2 \iff$

$2ab = a^2 + 2ab +b^2 -c^2 \iff$

$0 = a^2 + b^2 -c^2 \iff$

$c^2 = a^2 + b^2$

And that's that.

Of course we could have been more efficient. For example $([a+b] +c)([a+b] -c) = (a+b)^2 - c^2$ may have been more direct.

And in hindsight if we wanted to look like rock stars we could have done:

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

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

$\frac {ab}{a+b+c}=\frac {a+b-c}2$.