TITLE: Incircle Right Triangle Proof
QUESTION [3 upvotes]: 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.

REPLY [0 votes]: 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.
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