Trigonometry & Inverse Trigonometry
Properties of Triangle - Incentre
Grade 11
Question:
<p>97. In a △ABC, AD is the bisector of the angle A meeting BC at D. If I be the incentre of the triangle then AD : DI is equal to</p>
<p>(a) \((\sin B + \sin C) : \sin A\)</p>
<p>(b) \(2\cos A\)</p>
<p>(c) \(\dfrac{abc}{2\Delta}\cos A\)</p>
<p>(d) \(\dfrac{B-C}{2} : \cos\dfrac{B+C}{2}\)</p>
Step-by-Step Solution
Key Concept: The incenter I divides the angle bisector AD in a specific ratio determined by the sides of the triangle. Use the property that I divides AD such that AD/DI = (b+c)/a, where a, b, c are sides opposite to angles A, B, C respectively.
<p><strong>Step 1:</strong> Recall that the incenter I is the point where all three angle bisectors meet. AD is the angle bisector from A to side BC.</p><p><strong>Step 2:</strong> By the angle bisector property, D divides BC in the ratio AB:AC = c:b.</p><p><strong>Step 3:</strong> There is a standard result: If I is the incenter and AD is the angle bisector from vertex A meeting BC at D, then the incenter I divides AD in the ratio AI:ID = (b+c):a, where a = BC, b = AC, c = AB.</p><p><strong>Step 4:</strong> Therefore: AD/DI = AI/ID + 1 = (b+c)/a + 1 = (b+c+a)/a.</p><p><strong>Step 5:</strong> Since the semiperimeter s = (a+b+c)/2, we have: AD:DI = (b+c+a):a = 2s:a.</p><p><strong>Alternatively (Direct Form):</strong> AD:DI = (AB + AC + BC):(BC) = (b+c+a):a</p><p>∴ Answer: A (The ratio is <strong>(b+c+a):a</strong> or equivalently <strong>2s:a</strong> where s is the semiperimeter)</p>
Correct Answer: A