Let's denotes the semi-perimeter of a `DeltaABC` in which BC=a, CA=b and AB=c. If a circle touches the sides BC, CA, AB, at D, E, F, respectively. Prove that BD=s-b.

Explanation

A circle is inscribed in the `DeltaABC`, which touches the BC, CA and AB.

Given, BC=a, CA=b and AB=c
By using the property, tangents are drawn from an external point to the circle are equal in length.
`:.BD=BF=x` [say]
DC=CE=y [say] and AE=AF=z [say]
Now, `BC+CA+AB=a+b+c` `rArr(BO+DC)+(CE+EA)+(AF+FB)=a+b+c`
`rArr(x+y)+(y+z)+(z+x)=a+b+c`
`rArr2(x+y+z)=2s`
`[:.2s=a+b+c="perimeter of " DeltaABC]`
`rArrs=x+y+z`
`rArrx=s-(y+z)`
`rArrBD=s-b` `[:'b=AE+EC=z+y]`
Hence proved.