Trigonometry & Inverse Trigonometry
Angle bisector in triangles
Grade 11
Question:
<p>The lengths of the sides CB and CA of a triangle ABC are given by <i>a</i> and <i>b</i> and the angle <i>C</i> is \(\frac{2\pi}{3}\). The line CD bisects the angle <i>C</i> and meets AB at D. Then the length of CD is:</p>
<p>(a) \(\frac{1}{a + b}\)</p>
<p>(b) \(\frac{a^2 + b^2}{a + b}\)</p>
<p>(c) \(\frac{ab}{2(a + b)}\)</p>
<p>(d) \(\frac{ab}{a + b}\)</p>
Step-by-Step Solution
Key Concept: Use the angle bisector property combined with the area formula for triangles. Since CD bisects angle C, we can express CD using the areas of triangles ACD and BCD, which share the same height from their respective bases.
<p><strong>Step 1:</strong> Set up coordinates with C at origin. Let CA lie along a ray and CB along another ray with angle ∠ACB = 2π/3. We have CA = b and CB = a.</p><p><strong>Step 2:</strong> Since CD bisects angle C, it divides the angle 2π/3 into two angles of π/3 each. Point D lies on AB.</p><p><strong>Step 3:</strong> Use the area method. The area of triangle ABC can be expressed as the sum of areas of triangles ACD and BCD:</p><p>Area(ABC) = Area(ACD) + Area(BCD)</p><p><strong>Step 4:</strong> Express areas using the angle bisector property:</p><p>Area(ABC) = (1/2)·a·b·sin(2π/3) = (1/2)·a·b·(√3/2) = (ab√3)/4</p><p>Area(ACD) = (1/2)·b·CD·sin(π/3) = (1/2)·b·CD·(√3/2) = (b·CD·√3)/4</p><p>Area(BCD) = (1/2)·a·CD·sin(π/3) = (1/2)·a·CD·(√3/2) = (a·CD·√3)/4</p><p><strong>Step 5:</strong> Substitute into the area equation:</p><p>(ab√3)/4 = (b·CD·√3)/4 + (a·CD·√3)/4</p><p><strong>Step 6:</strong> Simplify by canceling (√3)/4 from all terms:</p><p>ab = b·CD + a·CD</p><p>ab = CD(a + b)</p><p><strong>Step 7:</strong> Solve for CD:</p><p>CD = ab/(a + b)</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D