Question

In a right triangle ABC,$\angle$ACB = 90$°$ a circle is inscribed in the triangle with radius r.
a,b,c are the lengths of the side BC,AC and AB respectively.Prove that 2r = a+b$-$c

Answer 1

As we know that,
$$\begin{gathered} \angle \mathrm{CPO}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \angle \mathrm{CRO}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \mathrm{In}\mathrm{quadrialteral}\mathrm{PCRO},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{PCR}+\angle \mathrm{CRO}+\angle \mathrm{ROP}+\angle \mathrm{OPC}=360°(\mathrm{Angle}\mathrm{addition}\mathrm{property}\mathrm{of}\mathrm{quadrilateral}) \\ \Rightarrow \angle \mathrm{ROP}=360°-\angle \mathrm{PCR}-\angle \mathrm{CRO}-\angle \mathrm{OPC} \\ \Rightarrow \angle \mathrm{ROP}=360°-90°-90-90°=90° \end{gathered}$$

Since all angles of $\square$PCRO are 90$°$, it is a rectangle.
So, we have:
CR = PO ...(1) (Opposite sides of a rectangle are equal)
CP = RO ...(2) (Opposite sides of a rectangle are equal)
CP = CR ...(3) (Tangents from a external point are equal)
OP = OR ...(4) (Radius of the circle)

From equations (1), (2), (3) and (4), we have:
CR = OR = OP = PC
We know that if all the sides of the rectangle are equal, then it is a square.
Hence, $\square$PCRO is a square.
So, CR = OR = OP = PC = r

$$\begin{gathered} \mathrm{Now},\mathrm{AP}=c-r \\ \mathrm{AP}=\mathrm{AQ}=c-r(\mathrm{Tangents}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{point}\mathrm{are}\mathrm{equal}) \\ \mathrm{BQ}=a-r \\ \mathrm{BQ}=\mathrm{BR}=a-r(\mathrm{Tangents}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{point}\mathrm{are}\mathrm{equal}) \\ \therefore a+b-c=\mathrm{BC}+\mathrm{AC}-\mathrm{AB} \\ =\mathrm{BR}+\mathrm{CR}+\mathrm{AP}+\mathrm{PC}-\mathrm{AQ}-\mathrm{BQ} \\ =a-r+r+c-r+r-c+r-a+r \\ =2r \end{gathered}$$