In a right-angled isosceles triangle ABC, ∠B is the right angle. The bisector of ∠BAC intersects BC at point D. If BD = 2 cm, then what is the length of CD?

Q.In a right-angled isosceles triangle ABC, ∠B is the right angle. The bisector of ∠BAC intersects BC at point D. If BD = 2 cm, then what is the length of CD?

Given: Since triangle ABC is a right-angled isosceles triangle, ∠BAC = ∠BCA = 45° Since AD is the bisector of ∠BAC, ∠BAD = 45°⁄2 = 22.5° Now, in triangle ABD: \[ \tan\left(\frac{45^\circ}{2}\right) = \frac{\sin 45^\circ}{1 + \cos 45^\circ} = \frac{\frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1 \] Therefore, \[ \frac{BD}{AB} = \tan\left(\frac{45^\circ}{2}\right) = \sqrt{2} - 1 \] So, \[ AB = \frac{BD}{\sqrt{2} - 1} = \frac{2}{\sqrt{2} - 1} = 2(\sqrt{2} + 1) \] Since triangle ABC is isosceles, \[ BC = AB = 2(\sqrt{2} + 1) \] Therefore, \[ CD = BC - BD = 2(\sqrt{2} + 1) - 2 = 2\sqrt{2} \] CD = \(2\sqrt{2}\) cm