Question

Let s denotes the semi – perimeter of a ∆ ABC, 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.

Answer 1

A circle is inscribed in the $\Delta$ ABC, 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 circles are equal in length.
$\therefore$ BD = BE = x
DC = CF = y
And AF = AE = z
Now, BC + CA + AB= a + b+ c
$\Rightarrow$ (BD + DC ) + ( CF + FA ) + (AE + EB) = a + b + c
$\Rightarrow$ (x+y) + ( y+z) + ( z+x) = a+b+c
$\Rightarrow$ 2 (x + y + z) = 2s
[ $\because$ 2s = a + b + c= perimeter of $\Delta$ ABC]
$\Rightarrow$ s= x + y + z
$\Rightarrow$ x= s - ( y + z ) [ $\because$ b = AE + EC = z + y ]
$\Rightarrow$ BD = s - b
Hence proved