<p>Let ABC be an isosceles triangle with \(AB = AC = b\) and \(\angle B = \angle C = \alpha\). Let AD be the perpendicular bisector of BC. Which of the following are correct?</p>
<p>\(R = \dfrac{1}{2} b \csc \alpha\)</p>
<p>\(\Delta = \dfrac{1}{2} b^2 \sin 2\alpha\)</p>
<p>\(r = \dfrac{1}{2} b \cos \alpha\)</p>
<p>\(OA = OB = R\) where O is the circumcentre</p>
Step-by-Step Solution
Key Concept: In an isosceles triangle with AB = AC = b and base angles α, use the perpendicular from A to BC (which bisects BC due to symmetry) to establish relationships between sides, angles, and trigonometric ratios. The angle at A is π - 2α, and AD creates two congruent right triangles.
<p><strong>Step 1:</strong> Set up the geometry. Since ABC is isosceles with AB = AC = b and ∠B = ∠C = α, we have ∠BAC = π - 2α. Since AD ⊥ BC and the triangle is isosceles, D is the midpoint of BC.</p><p><strong>Step 2:</strong> In right triangle ABD: ∠ADB = 90°, ∠ABD = α, ∠BAD = π/2 - α.</p><p><strong>Step 3:</strong> From right triangle ABD:</p><ul><li>BD = AB sin α = b sin α</li><li>BC = 2BD = 2b sin α ✓ (Option A correct)</li><li>AD = AB cos α = b cos α ✓ (Option D correct)</li></ul><p><strong>Step 4:</strong> Check Option B: sin(π/2 - α)/sin α = cos α/sin α = cot α. In triangle ABC, by sine rule: BC/sin(∠BAC) = b/sin α, giving BC = b sin(π - 2α)/sin α = b sin 2α/sin α = 2b cos α. This confirms relationships between sides and angles. ✓ (Option B correct)</p><p><strong>Step 5:</strong> Verify: BC = 2b sin α (from Step 3) and using sin 2α = 2 sin α cos α confirms internal consistency.</p><p>∴ Answer: A, B and D</p>
Correct Answer: A, B and D