Let \(s\) denote the semi-perimeter of a triangle ABC in which \(BC = a\), \(CA = b\), \(AB = c\). If a circle touches the sides BC, CA, AB at D, E, F respectively, prove that \(BD = s - b\).
\(BD = s - b\)
Step 1: Recall the property of an incircle. The circle inside a triangle that touches all three sides is called the incircle. From any vertex of the triangle, the lengths of the tangents drawn to the circle are equal.
Step 2: Mark equal tangents from each vertex. - From vertex A, the tangents are \(AF = AE\). - From vertex B, the tangents are \(BD = BF\). - From vertex C, the tangents are \(CD = CE\).
Step 3: Express each side in terms of tangents. - Side AB = \(AF + BF\). - Side BC = \(BD + DC\). - Side CA = \(CE + EA\).
Step 4: Let’s add these equations. \[ AB + BC + CA = (AF + BF) + (BD + DC) + (CE + EA) \]
Step 5: Rearrange using equal tangents. Notice that \(AF = AE\), \(BF = BD\), \(CE = CD\). Substituting, we get: \[ a + b + c = (AE + BD + CD) + (AE + BD + CD) \]
Step 6: Simplify. \[ a + b + c = 2(AE + BD + CD) \] So, \[ AE + BD + CD = \tfrac{1}{2}(a + b + c) = s \] (because \(s = \tfrac{a+b+c}{2}\)).
Step 7: Now focus on side CA = b. We know: \[ b = CE + EA \] But \(CE = CD\) and \(EA = AE\). So, \[ b = CD + AE \]
Step 8: Substitute back. From Step 6, we had: \[ AE + BD + CD = s \] Replace \(AE + CD\) with \(b\): \[ b + BD = s \]
Step 9: Rearrange. \[ BD = s - b \]
Final Result: Hence proved that \(BD = s - b\).