Given that △ABC is equilateral, and AD bisects ∠A, show that the side lengths of △ADB are in a ratio of #&{\sf 1}: \sqrt{\sf 3} : {\sf 2}#&.

✓ 1. Equilateral triangles are also equiangular, so
m∠A = m∠B = m∠C = 30° 45° 60° .

✓ 2. By definition of an angle bisector, m∠BAD = m∠CAD = 30° 45° 60°

✓ 3. Equilateral triangles are right obtuse isosceles triangles.

✓ 4. Because the bisector of the vertex angle of an isosceles triangle is the ⊥ bisector of the base, m∠BDA = m∠CDA = 30° 60° 90° .

✓ 5. BD = AB AC CD

✓ 6. 2( AB BD AC ) = BC

✓ 7. Triangles ADB and ADC are 30° - 60° - 90° triangles.
Let BD = 1. Then BC = 1 2 3 , which also means that AB = 2.

✓ 8. Using the Pythagorean theorem in △ADB,
1² + (AD)² = 2²
1 + (AD)² = 4
(AD)² = 3
AD = #&\sqrt{\sf 3}#&

Therefore △ADB is a 30° - 60° - 90° triangle with side lengths in a ratio of #&{\sf 1}: \sqrt{\sf 3} : {\sf 2}#&.