The given figure shows a right triangle right angled at B.

If ∠BCA = 2∠BAC, show that AC = 2BC.

Given: ∠BCA = 2∠BAC

To prove: AC = 2BC

Proof: In Δ ABC, sum of all angles is 180°.

⇒ ∠BCA + ∠BAC + ∠ABC = 180°

⇒ 2∠BAC + ∠BAC + 90° = 180°

⇒ 3∠BAC = 180° - 90°

⇒ 3∠BAC = 90°

⇒ ∠BAC = $\dfrac{90°}{3}$

⇒ ∠BAC = 30°

⇒ ∠BCA = 2 x 30°

= 60°

Using the trigonometry ratio,

sin 30° = $\dfrac{BC}{AC}$

⇒ $\dfrac{1}{2} = \dfrac{BC}{AC}$

⇒ BC = $\dfrac{1}{2}$ AC

Hence, AC = 2BC.