𝑎, 𝑏 and 𝑐 are the sides of a right triangle, where 𝑐 is the hypotenuse. A circle, of radius 𝑟, touches the sides of the triangle. Prove that  $r=\frac{a+b-c}{2}$.

We need to prove that  $r=\frac{a+b-c}{2}$ and it is given that  𝑎, 𝑏 and  𝑐 are the sides of a right triangle, where 𝑐 is the hypotenuse, and also, a circle, of radius 𝑟, touches the sides of the triangle. 

On drawing 𝑂𝑀⟂𝐶𝐵 and 𝑂𝑁⟂𝐴𝐶, we geth the figure as shown below.

Since, 𝑂𝑀⟂𝐶𝐵, 𝑂𝑁⟂𝐴𝐶 and 𝑂𝑀 = 𝑂𝑁 (radius), it can be said that 𝑂𝑀𝐶𝑁 is a square. 

Let 𝑂𝑀 = 𝑀𝐶 = 𝐶𝑁 = 𝑁𝑂 = 𝑟. 

It is known that the tangents from the external point to the circle are equal. Therefore, 

𝐴𝑁 = 𝐴𝑃, 𝐶𝑁 = 𝐶𝑀 and 𝐵𝑀 = 𝐵𝑃. 

Now, 𝐴𝑁 = 𝐴𝑃 can also be written as 

⇒ 𝐴𝐶 − 𝐶𝑁 = 𝐴𝐵 − 𝐵𝑃

⇒ 𝑏 − 𝑟 = 𝑐 − 𝐵𝑀 

⇒ 𝑏 − 𝑟 = 𝑐 − (𝑎 − 𝑟) 

⇒ 𝑏 − 𝑟 = 𝑐 − 𝑎 + 𝑟

 ⇒ 2𝑟 = 𝑎 + 𝑏 − c

$\Rightarrow 𝑟 = \frac{𝑎+𝑏-𝑐}{2}$

Hence, proved