Question

in the given figure ABC is a right triangle and right angled at B such that angle BCA = 2 of angle BAC. Show that hypotenuse AC = 2BC.
(hint ; produce CB to a point D that BC = BD).

Answer 1

GIVEN➡BCA=2BAC---------------1.)

In ∆ABC

➡angle BCA+ angle CBA + angle BAC = 180°

➡2 angle BAC + 90 + angle BAC = 180° [ from-1 ]

➡3 angle BAC = 90°

➡angle BAC = 90/3

➡angle BAC=30° and BCA=2×30=60°

Better use trigonometry

➡sin30=BC/AC

➡1/2=BC/AC

➡AC=2BC

(proved)

Answer 2

Using angle sum property of triangle ABC

∠ABC + ∠BAC + ∠ACB = 180°

⇒ 90° + ∠BAC + 2∠BAC = 180°

⇒ 3∠BAC = 90°

⇒∠BAC = 30°

∴ ∠ACB = 2 × 30° = 60°

Similarly, it can be proved that ∠ADB = 60° and ∠BAD = 30°

∴ ∠CAD = 30° + 30° = 60°

In ∆ACD, ∠CAD = ∠ACD = ∠ADC = 60°

So, ∆ABC is an equilateral triangle.

∴ AC = CD = AD

It is know that the perpendicular drawn to a side from opposite vertex bisects the side

∴ CD = 2BC

⇒ AC = 2BC [CD = AC]

Hence, proved.