Question 2 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.
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Solution
A circle is inscribed in the Δ 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. ∴ BD = BE = x DC = CF = y And AF = AE = z Now, BC + CA + AB= a + b+ c ⇒ (BD + DC ) + ( CF + FA ) + (AE + EB) = a + b + c ⇒ (x+y) + ( y+z) + ( z+x) = a+b+c ⇒ 2 (x + y + z) = 2s [ ∵ 2s = a + b + c= perimeter of Δ ABC] ⇒ s= x + y + z ⇒ x= s - ( y + z ) [ ∵ b = AE + EC = z + y ] ⇒ BD = s - b Hence proved