Question

In a triangle abc, angle b = to twice angle c, and the bisector of angle b intersects ac at d, then prove that BD/DA=BC/BA


Plz tell fast

Answer 1

AΔABC ∼ ΔABD.

Thus,

AB : AD = BC : BD

In ΔBDC,

∠DBC = ∠C

⇒DC = BD ...(2) (sides opposite to equal angles are equal)

Using (2) in (1), we get

AB : AD = BC : DC

⇒ DC : AD = BC : AB

Hope it helps..

Answer 2

In ΔBPC, we have

∠CBP = ∠BCP = y ⇒ BP = PC ... (1)

Now, in ΔABP and ΔDCP, we have

∠ABP = ∠DCP = y

AB = DC [Given]

and, BP = PC [Using (1)]

So, by SAS congruence criterion, we have

Δ ABP ≅ Δ DCP

Therefore

∠BAP = ∠ CDP = 2x and AP = DP ,

So in Δ APD, AP=DP

=> ∠ADP = ∠DAP = x

In ΔABD, we have

∠ADC = ∠ABD + BAD ⇒ 3x= 2y + x

⇒ x = y

In ΔABC, we have

∠A + ∠B + ∠C = 180°

⇒ 2x + 2y + y = 180°

⇒ 5x = 180°

⇒ x = 36°

Hence, ∠BAC = 2x = 72°