Question

∆ABC is a right triangle right angled at B such that ∠BCA = 2∠BAC. Show that AC = 2BC.

Answer 1

∆ABC is a right triangle right angled at B such that ∠BCA = 2∠BAC.

Produce CB to D such that BC = BD.

In ∆ABD and ∆ABC,

BD = BC (Construction)

∠ABD = ∠ABC (90º each)

AB = AB (Common)

∴∆ABD ≅ ∆ABC (SAS congruence axiom)

So, AD = AC .....(1) (CPCT)

∠BAD = ∠BAC .....(2) (CPCT)

Now,

∠BCA = 2∠BAC

⇒ ∠BCA = ∠BAC + ∠BAC

⇒ ∠BCA = ∠BAC + ∠BAD [Using (2)]

⇒ ∠BCA = ∠CAD

In ∆ACD,

∠DCA = ∠CAD (Proved above)

⇒ AD = CD (Sides opposite to equal angles in a triangle are equal)

⇒ AC = BC + BD [Using (1)]

⇒ AC = BC + BC (BC = BD)

⇒ AC = 2BC