Incircle Right Triangle Proof An incircle in the right triangle . The common point  of a circle and a hypotenuse divides the hypotenuse into two lines - the line  and . Prove that the product of the sizes of these lines is equal to the area of this triangle.

So far i used formula for incircle radius (r=(a+b-c)/2) to get the radius. next I created a square inside the circle with side = r. Then subtracted side of that square from side "a" and "b". translated these to sides into side "c" which now equals c = (a-r)+(b-r). I used this to get Area of the triangle S = (a-r)(b-r). And that's where i got stuck. I think i need to further transform this where the proof is clear that S = (a-r)(b-r) here is another picture for visualization of what I've done so far
thanks for any answers.
 A: You don’t need to find the radius of the incircle.
It’s enough to note that $a^2+b^2=(a-r+b-r)^2$
$a^2+b^2=a^2+b^2+4r^2+2ab-4ar-4br$
$0=4r^2-4ar-4br+2ab$
$2ab= 4r^2-4ar-4br+4ab$
$\frac{ab}{2}= r^2-ar-br+ab$
$\frac{ab}{2}= (a-r)(b-r)$
A: Since you have rendered $r=(a+b-c)/2$, simply substitute into the product $(a-r)(b-r)$:
$(a-r)(b-r)=\left(a-\dfrac{a+b-c}2\right)\left(b-\dfrac{a+b-c}2\right)$
$=\left(\dfrac{a-b+c}2\right)\left(\dfrac{-a+b+c}2\right)$
$=\left(\dfrac{c+(a-b)}2\right)\left(\dfrac{c-(a-b)}2\right)$
You then have $[c+(a-b)][c-(a-b)]=c^2-(a-b)^2=(c^2-a^2-b^2)+2ab$, where $c^2-a^2-b^2=0$ (why?) and thus your product reduces to $2ab/4=ab/2$. This matches the familiar "half the base times altitude" formula for the area, since each leg of a right triangle is the altitude to the other leg.

For a general triangle $\triangle ABC$: if the incircle divides side $\overline{AB}$ into pieces measuring $l$ and $m$, then the area of the triangle is
$S=lm\cot(\frac12\angle C).$
With a right angle at $C$ the argument of the cotangent is $45°$ so that factor reduces to $1$.
A: Hint: Proceed as follows from your diagram:
Call the center of the incircle $O$. Let the points where the incircle meets $AC$ and $CB$ be $D$ and $E$, respectively.
From here, we can easily figure out that quadrilateral $DCEO$ is a square.
We know that the length of any two tangents to a circle from any point are equal, so $AD = AT$.
We also know that $[ABC] = [AOB] + [ACBO] = [ATO] + [BTO] + [ADO] + [DCEO] + [BEO]$. If we call the length of the radius $r$, and use the formula that $r = \frac{a + b - c}{2}$, we can sum up the areas of each of these figures to prove our final answer.
A: Here is an answer that suggests that Point T is the point where the circle actually touches the line segment AB, and further that the line segment AT equals (a-r) and similarly that line segment TB is (b-r) in accord with the prior work courtesy of Tomi.
Proof for the Case of a 3, 4, 5 right triangle (as in this simply reduced case, the algebra is more facile).
The equation of the line AB is determined from its slope = -(0-a)/(b-0)= -a/b and the line passes through the point (b,0). So, its equation of line AB can be given by:

$$ {y - 0 = -a/b*(x - b) }$$

And, substituting a=3, b=4, we have the equation for the line for the special case of a 3, 4 and 5 right triangle as:

$$ {y = -3/4 x + 3 }$$

Note: the radius r as previously noted by Jack Smith has here conveniently the value r = (a+b-c/)2 = (3+4-5)/2 = 2/2 = 1. So, for the circle, the corresponding equation is therefore:

$$ {(x-1)^2 + (y-1)^2 = 1 }$$

The Point T is a claimed point of intersection which is given by intersection of this two equations. Upon substituting the provided value of y for the line AB we have:

$$ {(x-1)^2 + (-3/4 x + 3-1)^2 = 1 }$$

which can be shown to reduce to:

$$ {25/16x^2 - 5x + 4 = 0 }$$

which indicates that the value of the quadratic discriminant function is zero, implying a value of ${ x = 5/(2*25/16) = 8/5}$. And, per the above an associated value of ${ y= -3/4*8/5 + 3  = 9/5}$.
So, Point T has co-ordinates (8/5, 9/5) in the special case of a 3, 4, 5 right triangle.
Now, the y-intercept is at point (0,3) and distance between the y-intercept and Point T at (8/5,9/5) is the square root of:

$ { (0-8/5)^2 + (3-9/5)^2 = 64/25 + (6/5)^2 - 100/25 = 4 }$

So, its square root is 2 for line segment AT which is equal to a - r = 3 -1 or 2 as was required to be demonstrated.
In my opinion, a precise answer to this question requires three component: first, a source reference (or derivation) as to why r = (a+b-c)/2. Second,  the the algebraic derivation by Tomi and finally, the more general algebraic derivation for Point T and demonstration of respective line segment lengths.
