Question

ABC is a triangle in which ∠B = 2 ∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.
Prove that ∠BAC = 72°. 

Answer 1

In ΔABC,
∠B = 2∠C or, ∠B = 2y, where ∠C = y.

AD is the bisector of ∠BAC.
So, let ∠BAD = ∠CAD = x.

Let BP be the bisector of ∠ABC. Join PD.

In ΔBPC, we have
∠CBP = ∠BCP = y ⇒ BP = PC

In Δ′s ABP and DCP, we have
∠ABP = ∠DCP, 
∠ABP = ∠DCP = y

AB = DC     ........(Given) and,
BP = PC      ........(As proved above)

So, by SAS congruence criterion, we obtain

ΔABP ≅ ΔDCP

⇒ ∠BAP = ∠CDP and AP = DP

⇒ ∠CDP = 2x and ∠ADP = DAP = x ......[∴∠A = 2x]

In ΔABD, we have
∠ADC = ∠ABD + ∠BAD
⇒ x + 2x = 2y + x
⇒ x = y

In ΔABC, we have

∠A + ∠B + ∠C = 180^∘

⇒ 2x + 2y + y = 180^∘

⇒ 5x = 180^∘    ......[∵ x = y]

⇒ x = 36^∘

Hence, ∠BAC = 2x = 72^∘.